How to Calculate a Cube Root Using the Division Method

[@douglasfurman7163]

Dear Sir, Thank you for publishing this video on a cube root algorithm. I appreciate that you showed the algebraic justification too. I appreciate learning about historical mathematics. Best regards, Doug

[@CrystalClearMaths]

Greetings, Doug. Thank you for your feedback. I am glad that you have found my presentation useful. The algebraic justification is important to me (and to my students). It seems that we share a need, when learning something, to learn of its origins and 'raison d'être.' In other words, it is important to know WHY! Best regards to you, too! Graeme

[@dennisbell9639]

Graeme, I reviewed your presentation of calculating cube roots and now understand the method well enough that I even calculated the fifth root of a couple of rather large numbers.As you noted,the divisors and remainders do get rather large as you proceed through the method,making it not very practical above finding say a cube root.However,it is a lost skill and I'm very glad that you responded to that request and demostrated it for those of us who do appreciate the " old school " methods.Many moons ago when I was in grade school,I remember doing by rote the mechanical method of calculating square roots.Thanks to you,now it is finally crystal clear.

[@CrystalClearMaths]

+Dennis Bell Hello Dennis! It is lovely to hear from you again. You have certainly been "adventuring" with your root-finding lately! I am impressed that you tackled a couple of fifth roots. It is very satisfying to know that the method works and that you have succeeded in using it. Like you, I learned how to calculate square roots during my school days (actually, I was not taught the method at school but found out about it and learned it at home), but it was much later before I learned the principles of finding other roots by hand. I am delighted that you not only enjoyed the video but applied the principles to even greater problems :-) Thank you for letting me know.

[@awaviarybangalore]

I actually love maths but nowhere i found very useful content that teaches me about cubic roots, but when i switched to youtube all were about solving perfect cubes and i needed imperfect ones. So salute to you sir from India.👌

[@CrystalClearMaths]

Thank you, Abdul. I am glad that you found what you were looking for here. Kind regards to you from Australia :-).

[@cipherunity]

This is my first time to watch your presentation. I must say it is too good.

[@CrystalClearMaths]

I am glad that you found it useful and helpful, Afzaal. Thank you for letting me know.

[@gregorynelson1568]

I was playing around with the square root algorithm and sort of brute forced my way to this cube root algorithm by looking for patterns and testing, but I didn't know why it worked. It was brilliance when you showed and explained why it worked, that made so much sense now and I understand why the square root algorithm works now too. Thank you for this video.

[@CrystalClearMaths]

You are welcome, Gregory. I am glad that my explanation helped you understand the algorithms. Kind regards from Australia.

[@ishitarathi8907]

Couldn't find such an excellent explanation anywhere in school or youtube until i watched your video sir.... thank you so very much... i have an exam and this is very helpful.... recomending this video to everyone who are confused in finding the cube roots numbers

[@CrystalClearMaths]

Thank you very much, Ishita. I am glad that this video helped you (and thank you for recommending it).

[@WorldVibes]

That was an excellent video to explain what cube root is all about. You explained not only the way to calculate the cube root but also showed why that way is correct. I have seen videos where shortcuts are explained but nobody knows why they are correct. Thanks for making the concepts so clear to students watching this video.

[@CrystalClearMaths]

+Kamal Bhardwaj Thank you very much for your feedback, Kamal. I try to be thorough :-) Warm regards to you. Graeme

[@18rajdeepbakolia69]

was having difficulty in my chemistry exam for finding cube root, you helped a lot sir thank you:)

[@CrystalClearMaths]

I am glad that this video was a help to you, Rajdeep.

[@michaelknight8365]

Wow. This is such a great video. I love your explanation at the end showing how the process relates to (a+b)^3. Great work!

[@CrystalClearMaths]

Thank you very much, Michael. I am glad that you enjoyed the video and that you took the time to leave such an encouraging comment. Thank you.

[@ronalddump4061]

Btw, I am STILL the only person on earth (that I know of) who has demonstrated a practical method to find the cuberoot of any number using the long division method! Someone prove me wrong!

[@CrystalClearMaths]

You probably are, Ronald. I haven't forgotten about you ... just had a lot on my plate that has kept me from resuming work on the videos ... but I will get to your material eventually. I greatly appreciate your input and invite viewers of this video to follow up on our conversations in the comments below. Kind regards, Ronald.

[@ronalddump4061]

@@CrystalClearMaths Its good to hear"your "voice" again sir. Its about midnite here in the NW USA, sleep eludes me. Got up and turned on the PC. There you were. I never do this sort of thing, never eat at night, but here I am doing just that Hope you and yours are all doing well down there. Actually, It gets worse! If it appears that others do CR like I, that probably aint true. In reviewing the various threads under your vids, I am at least 4 people. The others are Archimedes Maid, Millicent Smallpenny, And the first, good ol' Rainman 37. You could say I am a "spit personality", but really I am quite normal, although too lonely. Need a good woman beside me. A few of my friends say that is an oxymoron, and that I am the lucky one. I think not. My algorithm is quite the standard long division method with some very obvious, very natural refinements. To figure it out, just concentrate only on what I did in the worksheet for the second digit, before you ever move on. I drew a border around the little worksheet for each iteration. Btw, in the meantime I demonstrated that I can evolve a cube root out to 25 digits on one side of one standard sheet. I think there is a picture in my phone. Sorry I am so cryptic and secretive, but my friends know that I love to taunt and poke fun at them. They also are sure I am a maths whiz. I am not really, it is just that they certainly are not. To the horror of my HS physics and maths teachers I didnt continue on to College. I remember one who was enraged. But I ramble. Hope you are having a beautiful day there. Thanx for remembering me

[@CrystalClearMaths]

@@ronalddump4061 It is 5 am here in Australia, Ronald. Little did I know we have communicated under different names 🙂. It is good hearing from you again. Thank you for giving me (and others) some insight into your approach to your algorithm. As I mentioned before, I look forward to getting to it as soon as I can clear some major jobs that have been occupying my time. I enjoyed your ramble and wish you a happy life! Kind regards from 'Down Under.'

[@CrystalClearMaths]

@@ronalddump4061 Hello, Ronald. I certainly remember you and thank you for your contributions. I thought I had replied to your last message (this one), but I revisited this page and found no message here. Perhaps it is a KZbin glitch? I hope you stay well and keep in touch. I have a very busy year developing. We shall see if I can manage to resume activity on KZbin or a similar site again. Thank you.

[@ronalddump4061]

Just checking in. I still don't know anyone else on earth who knows how to find ,CRs in an efficient manner Hope you are doing well down there. Winter coming on here soon Have been doing much salmon fishing Have a great summer

[@miguelreynel]

¡Excelente! Now I understand why, when extracting square roots, you have to multiply by 2 prior to look for the next number in the root, it’s the same logic. Thanks.

[@CrystalClearMaths]

Wonderful, Miguel! I am glad that the video helped explain this and greatly appreciate your taking the time to let me know. Thank you very much. Kind regards from Australia :-)

[@dennisbell9639]

I'm pressed for time,but would like to thank you for the demostrating the procedure for calculating the cube root of a number using an " old school " method.I'm going to spend the time practicing on some additional examples and will get back with you additional comments.I,too,enjoy the history of mathematics and the individuals who created it.Thanks again.

[@CrystalClearMaths]

+Dennis Bell Thank you for your encouraging feedback, Dennis. I am glad that you found the video useful and that you enjoy the history as well! I look forward to hearing how your practice goes. This is an interesting technique but it is time consuming. Best wishes to you (and nice to hear from you again). Graeme

[@gabrielgarcia1641]

I have not been able to find the estimate vs difference shown like this anywhere else. Well done.

[@CrystalClearMaths]

Thank you very much, Gabriel. I appreciate your encouragement. Kind regards from Australia!

[@Meghalova]

Thank you so very much for putting up this video. It really helped me to learn the long division method. Thank you so much.

[@CrystalClearMaths]

I am sorry that I missed your comment, Meghalova, hence this late response. I am glad that this video helped you :-) Kind regards, Graeme

[@gerulais]

I loved the binomial presentation. Until then I couldn't understand the whole multiplication with 30, 300 and so on. Thank you, sir.

[@CrystalClearMaths]

Thank you, Gerulais. On reflection, I could have made things even more clear, but I am still glad that it made sense to you eventually. Thank you for your feedback. Kind regards from Australia!

[@Transcentalpraetorian]

It's "Crystal Clear" sir thx alot!

[@TranquilSeaOfMath]

Thank you for posting this. I like how your algebraic explanation ties into the second course in algebra from the Ohio (U.S.A.) education standards. Best wishes. Cheerful Calculations!

[@CrystalClearMaths]

Thank you, TSoM, for your encouraging comment. I am glad that my algebraic explanation suits the standards in Ohio. I have always lived in Australia, but this skill is not taught in Australian schools. I learned it through extra reading and through applying algebraic and calculus techniques to the problem as I was always intrigued by how people in the past managed without calculators. So, the fact that my explanation fits with the standards in Ohio is a delight to me, but quite accidental/unintentional. Warm regards to you from 'Down Under.'

[@TranquilSeaOfMath]

@@CrystalClearMaths We definitely do not teach the application to cube roots. You would not likely find manual square roots taught either. We teach polynomial factors; and as you demonstrated they have applications. I try to discuss the geometry behind some of those 'facts'. Unfortunately students tend to see the factors as incomprehensible facts to memorize and not know how to apply 🙁. Keep teaching good mathematics!

[@CrystalClearMaths]

@@TranquilSeaOfMath Now I understand! Thank you, TSoM. I had noticed that most of my viewers seemed to be from the USA and assumed that this was a process taught in schools there. This makes much more sense. Thank you for enlightening me. Best wishes to you!

[@eddyhans5365]

COngratulations for your time explaning the root(..) of this calculation.

[@CrystalClearMaths]

Thank you very much, π.

[@tonybarfridge4369]

Your video came up again in the recommendations and I'd forgotten about them. My math has improved a little and I feel I understand it better now. No idea why I asked about decimals, but it looks like u did answer my question. However now I've forgotten how to get cube roots on the calculator so I can verify results. On a side note, one of my calculators is light charged and still works fine after 50 years and no battery replacement. It's the thin pocket type.

[@CrystalClearMaths]

Good to hear from you again, Tony! You need a scientific calculator for the calculation of cube roots. On my calculators, it is above the square root button ... so you would need to press [Shift][Square Root] to get it. Very best wishes to you.

[@tonybarfridge4369]

@@CrystalClearMaths I have scientific calculator apps but only one of them can do it yet it requires a lot of searching, trial and error. Best wishes to u also and I want to browse your videos now

[@CrystalClearMaths]

@@tonybarfridge4369 Thank you very much, Tony. I hope you find some material (among my videos) that you like. Very kind regards to you!

[@tonybarfridge4369]

@@CrystalClearMaths At the moment I'm focussing on cube roots; the long division and the shortcut. I'm trying to get my head around a 13 digit example I found somewhere based on the cubic algorithm. Also I've been interested in permutations and don't know if u cover those? Specifically regarding part perms of words with repeated letters.

[@CrystalClearMaths]

@@tonybarfridge4369 I don't recall that I have posted much about permuations, yet, Tony. They are on the agenda when I resume posting videos. Sorry that I can't help you at this stage.

[@malazeshafurfur5291]

I find ur video very useful sir. Thank u very much for making this video.

[@CrystalClearMaths]

You are very welcome, Zesha. All the best for your imminent exam (and thank you for subscribing). Warm regards, Graeme

[@DarthVader22576]

Nice Work. I was looking for a video like this which could algebraically explain the algorithm of calculating the roots of a rational number. Thank You. 😁

[@CrystalClearMaths]

You are welcome, Vivaswan.

[@robertfindlay2325]

Thank you. I was trying to figure out how to calculate logarithms, and I think hat if one can calculate square and cube roots one can use these to calculate logs to base 10 using paper and pencil. Problem was i couldn't quite figure out how to calculate cube roots.

[@CrystalClearMaths]

Roots and logarithms are closely linked, of course ... so you are exploring a rich mathematical topic. Let me recommend Eli Maor's book, "e: the Story of a Number." I think you will enjoy some of the insights that he provides. Best wishes to you, Graeme

[@cristianoronaldo3875]

Hello sir. It's been a long time and i still watch your videos and remember you. I really hope that you start making videos again... Wish you a very healthy and happy life Warm regards~

[@CrystalClearMaths]

Thank you very much, friend. I am glad that you still find value in these videos. That really encourages me. Thank you. At this stage I expect to resume creating and posting videos around the middle of the year (June or July). There are some other matters that I have to deal with in the meantime. I wish you a healthy and happy life, too. Grace and peace to you.

[@fatcat5602]

@@CrystalClearMaths every time I try doing cube root of 300 (which is approximately 6.694), I get 84 as the difference between 300 and the cube of the first digit (216). But when I do 6^2 * 300, which brings out 10800, and divide 84000 by 10800 I get approximately 7.778, when the approximation should bring out 6 as the second digit. What am I doing wrong here?

[@ronalddump4061]

@@fatcat5602 The best way to do this is after you determine the first digit 6, cube it and subtract it and get 84, just as you did. Bring down the next 3 digits 000 = 84000 Now on the worksheet square that 6 and multiply by 300 = 10800. beside that multiply the 6 by 30 = 180. These 2 numbers together give your next root digit. Obviously 8 cannot work, because 8 times 10800 alone gives more than the 10800. But can 7 work?? When you have estimated the next digit, add it to the 180 = 187. Multiply that by the 7 = 1309. Add that to the 10800 = 12109. Multiply that by the 7 = 84763. Still wont work. So the second digit is 6. 186 x 6 = 1116. Add that to the 10800 = 11916. 11916 x 6 = 71496. Put that in the LD problem under the 84000 and subtract. Bring down the next 3 digits, 000, and begin the next iteration doing the same as above. With all due respect to him, this is the very best way. 300a^2 and 30a on your worksheet, and that is all. The next digit estimation division can usually be entirely done in your head.

[@archimedesmaid3602]

@@fatcat5602 You are not realizing that the total divisor is not just a^2 x 300. It is actually a combo of a^2 x 300, and a x 30. When you estimate your next digit (which you erroneously estimated as 7, you prepare by squaring your "a" which is 6, and multiplying by 300. You get your 10,800. (Which you did!). Now you also multiply a by 30 = 180. You then add your 7 to 180 = 187. And multiply that 187 by the seven = 1309. Add that to your 10800 = 12109, and multiply by your 7 = 84,763. That is bigger than 82,000 which was your remainder, so the second digit actually is 6 I realize I showed you a different process than he did. But my method is much cleaner and concise, while his is something of a cluster f**k, and contains redundant operations. (Notice how he actually does 300a^2, but only as a method to estimate the next digit. Then he actually erases that answer! This is foolishness! ). If you apply yourself you can find the CR of any number to 7 digit accuracy on about 1/4th of a standard sheet of paper, and take only about 10 minutes. If he tried that using his process it would take at least an hour or two and use many whole sheets of paper What is the CR of 470 215.567 832 992 131 to 7 digit accuracy? You can learn to do that in less than 10 minutes every time.

[@bgm1140]

@@archimedesmaid3602 Thank you very much for your advice!

[@jimcurrie]

I like your method. My high school teacher was a little bit more abstract on cube roots

[@CrystalClearMaths]

Thank you Jim. I am glad that you found this useful and comprehensive (comprehensible). Kind regards to you from Australia.

[@purvauppal1640]

Thank you sir for explaining. This really helped me a lot...

[@CrystalClearMaths]

You are welcome, Jyoti. I am glad that I was able to help you via this video :-) Kind regards.

[@martinryan9141]

Hi Sir, I am so pleased to have found this video which is the best explanation that I have come across after 3 days of searching on the internet. I have an old maths book which was handed down by my grand-aunts who were teachers in the early part of the 20th century. The book does not show when it was published but I suspect it was around the 1900s. There are a couple of examples of how to do the long calculation but I did not manage to follow the logic. I learned how to do cube roots in the very early 1960s (before calculators!) and was good at maths in general.I am now 70 and need to keep the remaining few brain cells ticking over....don't ask me why. By the way, I had difficulty with trigonometry when at school but managed to learn a quite a bit of it in the past year. If I keep going I might even sit for a degree, only joking. I notice from one of your replies that you were having some medical issue so I hope you are now back to full health. Just in case you want to check up on the maths book; it was published by G.H. Bradford, London under series titled ' Jack's Mathematical Series, Intermediate Arithmetic. Regards and Thank You, martinjr.

[@CrystalClearMaths]

Thank you, Martin. Your kind feedback is greatly appreciated! It is particularly encouraging to find someone continuing to use and enjoy their mathematics skills in later years (I am in my sixties). Given what you have shared, you may also find some of my trigonometry videos useful (kzbin.info/www/bejne/bWrSaoGYocedo68). I was intrigued by your mathematics book, having not heard of it before. It must be exceedingly rare today because I could find no mention of it on www.bookfinder.com and no detailed information via a general Google search. It is good that you are able to enjoy one of those earlier texts. I have a number of older text books in my library and use them to my great benefit. My health issue is a chronic one (without going into details). It is not life-threatening, but is definitely life-changing, and severely limits my production of videos etc. Sorry to be so mysterious, but I prefer not to give too many details in such a public forum. Thank you for your concern and warm wishes. Please keep in touch. I hope you continue to enjoy my videos and find them useful. Warm regards, Graeme

[@martinryan9141]

Hi again Graeme. I gave you some wrong info. G H Bradford was the writer and the publishers were TC & EC Jack, Henrietta Street, London. One of the questions in the book refers to a Promissory Note with the date July 1st 1902.If you pm me I can send an email with scanned copies of the cover, the page with the title and publishing details and the page with the long version of a method to calculate the cube root - the one I studied but couldn't figure out. I do now after seeing your video.The sample manipulates the figures in a slightly different way to yours but with the same end result. Let me know if you would like to have the scans, plz. Regards, Martin

[@CrystalClearMaths]

Hi Martin, I understood that Bradford was the author and am quite intrigued by your description of the book. The scans would be greatly appreciated, thank you. I cannot see how to contact you privately via KZbin as your permissions are set rather tightly. Perhaps you could PM me via KZbin or send me a message via my website (www.crystalclearmaths.com) and we will take it from there! Thank you again. Warm regards, Graeme

[@MrVoayer]

Dear Graeme, this was an impressive presentation. Detailed and with a lot of information and hints. Only, it will take me several jugs of coffee to get it all in ;-)Good job, nonetheless!

[@CrystalClearMaths]

+MrVoayer Thank you very much, MrVoayer. It is an involved technique! It helps me to appreciate the skill and dedication of mathematicians in the past who used methods such as these regularly. Enjoy your mix of coffee and mathematics :-) Warm regards to you, Graeme

[@jonathan5156]

Bro made this more simple than my teacher ever could

[@CrystalClearMaths]

Thank you, Jonathan. Obviously, the video helped you and for that I am glad. 🙂

[@prakashs5155]

Thanks sir I learned a easy way to solve cube root of any number sir thanks sir 😀😁😂🤓🤓

[@CrystalClearMaths]

I am glad that this video helped you, Selva. Best wishes for your studies.

[@saumitraanand286]

Thank u very much sir, for this sure shot method...

[@CrystalClearMaths]

You are welcome, Saumitra.

[@guitarttimman]

I'll answer your question as to why someone might want to learn this method. It is because it stimulates the mind and helps one to become more intelligent. It also sharpens one's multiplication, addition, and subtraction skills.

[@CrystalClearMaths]

Point taken, Tim. You are quite correct. Thank you.

[@destruidor3003]

Extremely good video….you shared a good piece of knowledge…..great work

[@CrystalClearMaths]

Thank you, Destruidor.

[@vaishaliahirrao777]

Sir how we solve for trigonometry functions whole square root or cube root?? Like (sin ^-1 x)^2 ?? Sir please explain

[@CrystalClearMaths]

I do not understand, friend. I am sorry. Perhaps you could give me a few examples. I will not reply for a while. I need to sleep.

[@What_A_Great_Dude]

Thank you for the video sir but how do I find the cube root of one, two digit, four or more digit numbers.

[@CrystalClearMaths]

Remember to collect the digits into groups of three, Great Dude 🙂 ... can I call you 'Great' for short? So, to find the cube root of 16, treat it as 016. To find the cube root of 5, treat it as 005. To find the cube root of 1784, treat it as 001 784. If you need decimal places added, you keep adding zeros in groups of three after the decimal point. E.g. 001 784.000 000 000 000. I hope that helps. The process is the same is in the video. Thank you for your question.

[@mathtutoronline2088]

Amazing and accurate information explaining

[@CrystalClearMaths]

Thank you, kindly, MTO.

[@JasonCoffman-xu5ks]

​​​@@CrystalClearMaths sir when you multiply the 8^2 by 3 and then add 2 zeros, (300x^2) you suggested that this was just to produce an estimated next digit. If you do that you are really throwing away work unnecessarily. Simply multiply that 300 by 8^2 , then also multiply the 8 by 30. Use this, just as you did, to determine your next.digit. Add that 5. to the 30x figure. Multiply that by the same 5, add that to the 300x^2 figure and multiply by the 5 again. And you are done. Put that under you current remainder and subtract. Bring down the 000 and you have your new remainder. Continue. You are unnecessarily discarding important work which you have already accomplished. The 300x^2 number is in no way a "scribble" throwaway number as you suggested This perfectly satisfied the a^3 + 3a^2b + 3ab^2 + b^3 formula in a much more concise and understandable way😂

[@JasonCoffman-xu5ks]

​@@CrystalClearMaths ps, this would have been the ancient way to make charts of cube roots. And they also would have had more nifty methods to vastly abbreviate the process if they wanted to make charts of cube roots out to possibly 5 digits of accuracy

[@CrystalClearMaths]

@@JasonCoffman-xu5ksThank you very much for your astute and helpful contribution, Jason. It is appreciated. I hope others will also benefit from what you share.

[@CrystalClearMaths]

@@JasonCoffman-xu5ksI dare say that you are correct, Jason. I admire those who managed to calculate these tables and make them available to others. It was a huge task.

[@yt-bj8fi]

Excellent presentation. Thank you!

[@CrystalClearMaths]

Thank you, yt. (And you are welcome.)

[@shivampurohit1331]

Dude, your vids are great! Continue on! Gods bless you

[@CrystalClearMaths]

Thank you very much, Shivam. I am delighted to learn that my videos are helping people (like you). Your feedback is very encouraging. Best wishes for your studies and mathematical explorations.

[@dariuseversol]

here are the formulas that I am looking for the one which is correct or more correct rather : number 1:) estimate = ((n-1) * Old estimate + M/(old estimate^(n-1))/n or Number 2 : ) estimate = (old estimate+(n-1)*(m/(old estimate^(n-1)))/n which is most correct? and can you show the correct veresion of this formulah please I saw it in a book and it has been a while and I forgot it... I do know that it can converge very quickly onto the nth root of a number...

[@CrystalClearMaths]

Thank you, Darius. I hope you see my other comment in response to you. I have kept this comment also and will try to reply when I can. Thank you very much for your interest and patience. Best wishes to you.

[@hussainhussain6963]

Uncle you are great thanks for vedio

[@CrystalClearMaths]

You are welcome, Hussain. Best wishes for your studies!

[@rajendrarajak5649]

Thank you sir your video helped me a lot

[@CrystalClearMaths]

You're welcome, Rajendra. I am glad that the video helped you. Thank you for letting me know.

[@theworldofcreativity9071]

wait what happens if in the calculating the difference part we get a number>10?

[@CrystalClearMaths]

Occasionally, that happens. It is simply an estimate. In that case, choose 9 and see if it 'works.' This is not an exact predictive method. It is an iterative process using approximations. Thank you for seeking clarity, WoC. Sometimes, the first approximation needs to be adjusted.

[@PankajSoni-hm4vd]

Sir if 0 is there in place of 5

[@CrystalClearMaths]

Hello Pankaj. Thank you very much for your comment. Unfortunately, I do not understand what you mean. I am sorry.

[@PankajSoni-hm4vd]

Sir I am asking that if we have choose some no like 29 then the second no will come 0 so how we will do in that case then multiply by 0 will be 0

[@CrystalClearMaths]

​@@PankajSoni-hm4vd You will start with 029.000 000 and your first estimate would be 3 because 3³ = 27, which is just a little less than 29. Subtracting this from 29 leaves 2 which we write underneath. We then draw the next three zeros down to make 2000. Our estimate for the next digit (the first decimal place of the cube root of 29 is calculated by finding the estimate on the left first ... which will be 3² x 300 = 2700. 2700 does not even go once into 2000, so our first decimal digit will be zero. This means that we will draw down three more zeros to make 2,000,000. Our estimate on the left will now be 30 (from our estimate of 3.0) squared x 300 = 270,000. This divides into 2,000,000 just over 7 times, so our estimate for the second decimal place is 7 (and we have 3.07 as the approximate cube root of 29. Of course, we will have to confirm the 7 with our careful calculations on the right. I hope this helps, Pankaj. Best wishes to you.

[@PankajSoni-hm4vd]

Sir I forgot to estimate with 30 Thanks Sir

[@CrystalClearMaths]

@@PankajSoni-hm4vd You are welcome, Pankaj :-)

[@taran7954]

Uploaded in 2015! Ah, man the good ol' times..

[@CrystalClearMaths]

;-)

[@alana7995]

You have no idea , it's 2020 now

[@ms74510]

@@alana7995 2021 now xd

[@satyajitkar9787]

your videos are helpful thank you sir

[@CrystalClearMaths]

You are very welcome, Shubhadeep. Thank you for letting me know.

[@tonybarfridge4369]

I thought I had a handle on these Graeme, yet now there's some problem. For CR 17 I get 2.7, when the answer is 2.57. To start off I do 2^2x300, and divide into 9000 and get 7 every time. By doing the next part and using the 7 as "b" I get the correct decimal of .5 but the next one produces 3 when it should be 7.

[@archimedesmaid3602]

That is because there is more involved in determining the divisor. There is 300a^2, as you suggest, but also, besides that, there is 30a Remember, his 300a^2 = 1200 was only a TRIAL devisor With all due respect to Graeme, there are simpler, more clear ways to use his algorithm, and they end up with the same results Here: You used 300a^2 as your trial divisor. Just as he does. But if you use your 7 as your nexxt digit, you will come up with 11683 to subtract from your remainder of 9000. Obviously that is WAY too large. Try 6 as the second digit. [1200 + ((30 x 2) +6) x 6] x 6 = 9576. That is still larger than your current remainder of 9000, so 6 cannot be the 2nd digit So 5 must be the second digit. Which gives you 7625 to subtract from your currant remainder of 9000. So now your next current remainder is 1,375,000. Proceed as before Look how much simpler/clearer that was than his method. You should not be thinking that the 300a^2 thing is only to be used as a trial devisor

[@tonybarfridge4369]

@@archimedesmaid3602 Thanks so much. I'm trying to do these for personal interest only, but realised I still can't. When I have time I will try to work it out according to your suggestions. When he said it was only a trial I misunderstood, and thought the right column was only to verify that result.

[@archimedesmaid3602]

@@tonybarfridge4369 Remember, in each iteration the first thing you do in your workspace is square your current root "a" and multiply by 300. Beside that multiply a by 30. So 300a^2 and 30a. (NOTHING ELSE -- this is all you will need). When mentally considering your trial divisor you incorporate both of these. (Btw, eventually your "trial digit" will be not so much trial, you will get it right the first time 95% of the trials). In each iteration, after you determine your trial digit "b", you : #1 -- Simply add that b to that 30a figure, #2 -- multiply that by b, #3 -- add that to the 300a^2 figure, #4 -- and multiply again by b. And you are done! #5 - Put that under the current remainder in your LD problem, subtract, bring down the next group of 3, and proceed with the next iteration. Done! look how simple that was! Notice this is much different than what he does. FI, he does the 300a^2 thing only to determine his b, then oddly erases it. Dont do that! It is horrible redundancy! That 300a^2 is the MAIN part of your problem. Dont ever erase or discard things in your workspace! The purpose of his vid was only to show you why and how this algorithm works and he did a great job of it, by using the a^3 + 3a^2b + 3ab^2 + b^3 formula. But in practice my algorithm does the same thing much, much tidier. My algorithm exactly satisfies that formula In the same way, for sqrts, simply, in your workspace multiply a by 20. Add your b to the 20a figure. Multiply that 20a + b figure by b. Done! Very simple, VERY easily remembered forever! (Just as the 300a^2 and 30a thing is very easily remembered forever), unlike the popular sqrt method taught on some utube vids. These 2 procedures for CR and sqrt are certainly the ancient methods for finding roots.

[@archimedesmaid3602]

@@tonybarfridge4369 Btw, it is special and sweet to watch your "childish" fascination with numbers! Dont ever lose this. It will help keep your mind young and agile into very old age

[@tonybarfridge4369]

@@archimedesmaid3602 Many people including mathematicians have a fascination with numbers. Never heard it described like that, but thanks anyway. I've learned a lot online but also from developing my own methods which is even more satisfying

[@Muck-qy2oo]

You can also use the Newton-Raphson, or the Herons method!

[@CrystalClearMaths]

You are quite correct, Muck2014. Thank you. I should have pointed out that there were other methods that could be used. I chose to explain and demonstrate this (rather cumbersome) method because it does not require my explaining any calculus (Newton-Raphson method) and because it appears to follow the approach used in US schools. I say this as a distant observer from Australia. Interestingly, the method I used in the video is not taught at all in the schools here, but students do learn to use the Newton-Raphson method in their Extension 1 course (during their last two years of High School). As far as I know, no Australian school introduces Heron's method. Thank you for your observation ... and kind regards.

[@Muck-qy2oo]

@@CrystalClearMaths Yes, this method actually isn't really taught in any school in the world as far as I know. Even though it's still quite simple! Herons method doesn't need to be looked at as difficult calculus in my opinion. I learned it just by experimenting with computing averages. You can extend it easily even for cube roots of course.

[@CrystalClearMaths]

@@Muck-qy2oo Thank you, Muck2014. I had been under the (probably mistaken) impression that this kind of method was taught in US schools. Having calculated square roots manually as a child (learned outside the school system), thought it would be fun/instructive to make this video to develop the idea a little. I had heard of Heron's Method but not actually used it myself. It looks quite interesting and, as you say, not terribly difficult. Hopefully, some viewers will read this conversation and be inspired to explore further :-). I greatly value your input ... as well as your taking the time to post your messages. Thank you!

[@archimedesmaid3602]

@@Muck-qy2oo If you apply yourself you can find the CR of any number to seven digit accuracy (using a form of this LD method), on a tiny scrap of paper, in less than 10 minutes! Pen/paper only, no calculators involved befor or during Only very simple addition, subtraction, and multiplication, and a bit of mental division to estimate the next digit. (And you will never at any time ever need to multiply a larger number by any numbers other than 2, 3, 4, 6, 8, 10, 12, 14, 16, and 18.) Can you do that with your method?

[@Muck-qy2oo]

@@archimedesmaid3602 I only need simple division and some addition and multiplication for Herons Method. It doesn't increase as much in its computational effort with higher powers as the LD method does. The LD method becomes quite nasty with higher powers. I once tried to compute the 5. and the 7. and 11. root of numbers with this and it took about 30 minutes just to get the answer to 3 digits, with all the errors and paper writing work it takes for the compuatational effort. So no I don't see how one can use this LD method as easily as it is for square roots on higher powers. Of course the LD method which is based on the binomial theorem works exactly digit by digit while calculus only works approximately. But Herons method converges quadratically. As an example: the cube root of 50 is guessed as 3,6. Now you have to divide 50 by 3,6² which gives 3,85802469 Than you do (3,85802469+3,6*2)/3 which is 3,686068 For the second iteration you will get 50/3,686² = 3,680 Now after the second iteration we get (3,680+3,686*2)/3 = 3,684 For the third iteration we take 50/3,684² = 3,684094496 Iterate for (3,684094496+3,684*2)/3 = 3,6840315 After 3 iterations we got an accuracy to 7 digits with a 3 minute computation and and an error of 1,107*10⁻⁷ %. With the binomial long division method I would have gotten only 3 accurate digits and if I make a mistake I will be off without noticing it first. 623 is guessed as 8 because 8³ is 64*8 = 480+32 = 512 1. 623/8² = 9,734 (9,734375+2*8)/3 = 8,578125 2. 623/8,578125² = 8,466488 (8,466488+8,578125*2)/3 = 8,54091266 3. 623/8,54091266² = 8,540425 (8,540425+2*8,54091)/3 = which is 8,54074833 and that's accurate to 99,99998 %! The computer gives me 8,54075012 The answer cubed gives as cubed 622,9996 and that's 99,9999 % accurate!

[@laytonjames9741]

Apologies if this has already been addressed. During your algebraic discussion you say that a=80 and b=5, but then you say the a^3=512; is a=80 or a=8 ?

[@CrystalClearMaths]

I am sorry, Layton. I was being a bit lazy there, and should have been more careful. The value of a = 80 as stated. When I wrote (a + b)², the 623 underneath represented the 623 000 from the division (using the second group of three digits ... all zeros). The 512 that I wrote under the a³ should have been 512 000, and the difference that we calculated underneath is our attempt to check that the second digit, 5, actually does get us very close to 623 000. I should have been much more careful and appreciate your pointing this inconsistency out. Thank you.

[@arshantv3579]

sir plz tell us the method of finding cube root of 2 3 4 5 6 etc

[@CrystalClearMaths]

Hello Arshan. Unfortunately, all I can suggest is that you try to follow the steps in this video and calculate them for yourself. It is not an easy process, but you will feel better for having used it once or twice. You will certainly appreciate the privilege of having and using an electronic calculator! Kind regards.

[@arshantv3579]

@@CrystalClearMaths thank a lot sir may almighty lord bless u

[@CrystalClearMaths]

@@arshantv3579 May the Almighty Lord bless you, too, Arshan.

[@dariuseversol]

can you please show the method of finding nth roots by taking an estimate and then raising that to N-1 and dividing and then taking either n-1 times the estimate plus the result or n-1 times the result and adding to the estimate and dividing by N to get a new estimate etc?? because this method can diverge very quickly to the answer and in my mind is very elegant and beautiful but I forgot the exact "correct" formulah thanks

[@CrystalClearMaths]

Hello, Darius. I am sorry that I have taken so long to reply. I have been very preoccupied with other matters for some time and have been unable to devote serious time to KZbin videos or comments. I am keeping your comments and hope that I can respond to you in a much more full and complete way in a few months. Please accept my apologies for the delay. Kind regards from Australia.

[@aasthatoppo2412]

thanku sir..this video was really helpful

[@CrystalClearMaths]

I am glad that it helped you, Aastha. Thank you for letting me know. Best wishes for your studies.

[@SalmanShaikh-zt7uz]

Can you please find the cube root of 350 by same method, I'm confused how to find a cube root of 350 by long division method

[@CrystalClearMaths]

Yes, Salman. This method may be used to find the cube root of any number. Some estimates will prove more easy than others, of course. Watch the video again and keep trying. You may have to alter your original estimates occasionally. Kind regards to you ...

[@arshantv3579]

sir in case of cube root of 62 than the first estimate is 3^2×300 or 3^2×30 if we take 3^2× 300 = 2700 the value of cube root of 62 is = 3.1 but in calculater the value cube root of 62 is 3.9

[@CrystalClearMaths]

Hello Arshan. When we estimate the first digit to be 3, we subtract 3³ from 62 to get 35. When we draw the three 0s down, we get 35,000. Our 'estimate,' as you correctly say, is 3² x 300 = 2700. Dividing 2700 into 35,000 gives us a number in excess of 10, so we would make our estimate for the first decimal place to be 9 ... i.e. our estimate now becomes 3.9 We confirm this by completing the more rigorous calculations in the right-hand column. I hope this clarifies things for you. Nice to chat with you again :-)

[@arshantv3579]

cube root 62 =3.957 i have got up to 3.95 but found difficulty in third in place 3.95^2×300 =456300 estimate now we have divide to 745375000 acc to calculater third place decimal is 7 secondly if we take 7as a 3digit no a= 30 b=957 =(3a(a+b)+b^2)b =(3×30(30+957)+957^2)957 =(90×987+915849)957 =(88830+915849)957 =1004679×957 =961477808 its horible dont u think my sir

[@CrystalClearMaths]

@@arshantv3579 Hello Arshan. I can see that I did not make my video very clear at one point. When you have finished calculating the difference after reaching 3.95, you should have had a difference of 2,681,000 - 2,310,875 = 370,125 to which you add 000 to begin the next round. So, your target number will be 370,125,000. Your estimate will now be 395² x 300 = 46,807,500. Dividing 370,125,000 by 46,807,500 will give you 7.907 386 637, so 7 looks like a good estimate for the next decimal place. What I may not have made clear is that, at this stage, a = 395 and b = 7. In other words, a is the value that you have already confirmed (without the decimal point) and b is the extra digit that you wish to add (after you have tested its accuracy). The a + b probably should read 10a + b since you are to simply use your full estimate at that location. This means that your estimate is [30a(a + b) + b²]b = [30 x 395(3957) + 7²]7 = 328,233,493 Subtracting this from your target number gives 370,125,000 - 328,233,493 = 41,891,507 which is less than your estimate of 46,807,500, so it appears that 3.957 is your best estimate for ³√62 so far. I apologise for not making the process as clear as it should be. If I re-recorded the video, I can see some improvements that I would make. Nevertheless, I hope that this explanation helps clarify things for you. Kind regards.

[@arshantv3579]

hmmm good sir ihope the next time u make a new video part 2of cube root in which u will show how to find a cube root up to 3....4 place of decimal

[@CrystalClearMaths]

@@arshantv3579 I agree that I should refine this video. It is on the list!

[@gabigolf5752]

How could you do the cube root of 178? I keep getting my first estimate wrong!

[@CrystalClearMaths]

Hi Gabi. Sadly, I do not have time at the moment to work these out by hand. You would first estimate 5 and subtract its cube (125) from 178, giving 53. Drawing down the three zeros means that our new target is 53,000. On the left, your estimate squared x 300 would give 5 x 5 x 300 = 7500. Dividing this into 53,000 gives 7 (and a tiny bit left over), suggesting that 7 should be the next digit. When you go through the procedure on the right, you will realise that 7 is a little too big, so you would have to then revert to using 6 as your second digit (i.e. 5.6) and you would find that it works! This method will occasionally throw up incorrect digits as you are always using estimates. On these occasions, you will have to back-track a little and try the next digit. Kind regards, Graeme

[@tonybarfridge4369]

You mean (3a(a+b)+b^2)b is the same as the cube formula?

[@CrystalClearMaths]

Basically, yes, Tony. If we want c (the cube root of c³) and we know that a is a close approximation to that root, then a³ is very close to c³. The difference between them can be calculated in this way ... let c = a + b (i.e., let the small difference between your estimate and the 'real' value of the root be represented by 'b'). Then c³ = (a + b)³ = a³ + 3a²b + 3ab² + b³ Then c³ - a³ = 3a²b + 3ab² + b³ = [3a(a + b) + b²]b which is the difference that you were asking about. Kind regards to you, and thank you for your question.

[@tonybarfridge4369]

​@@CrystalClearMaths Wow u kinda lost me there but I'm working on the understanding. What I was asking was if that second formula (?) was the same as the original cube formula. Or is it an approximation of it? I wondered if it was a simplified version which still gave the exact same result? Probably u answered that tho. I found the video easy enough to follow until u started solving for decimals. Also I will check out your others. Thanks very much for the response.

[@CrystalClearMaths]

@@tonybarfridge4369 I am sorry that my explanation was confusing, Tony. I probably did not understand your question properly. Solving for decimals is conducted using the same algorithm/method. If you began with a very large number, you will obtain the same numerals as you would if the number had a decimal place relocated by a multiple of three digits. For example, using WolframAlpha ... cuberoot(43 859 327 355) = 3526.5820199270323025580200590609934244100412403365115626594 cuberoot(43 859 327.355) = 352.65820199270323025580200590609934244100412403365115626594 cuberoot(43 859.327 355) = 35.265820199270323025580200590609934244100412403365115626594 cuberoot(43.859 327 355) = 3.5265820199270323025580200590609934244100412403365115626594 cuberoot(0.043 859 327 355) = 0.35265820199270323025580200590609934244100412403365115626594 The shift of three digits in the original number is needed because the cube root of 1000 is 10. This results in a decimal shift of one digit in the answer. Apart from that, there is no difference in method between calculating numerals before a decimal point or after one. I hope you find some of the other videos interesting. Warm wishes to you from Australia :-)

[@tonybarfridge4369]

@@CrystalClearMaths Thanks man, I knew about the cube root digits from sets of 3, anyway I'm an Aussie too😁

[@anonymous4evr746]

hey can u show me how to find the cube root of 25.454 using the same method.....

[@CrystalClearMaths]

I will try to get to it, Anonymous. Unfortunately, I have not been well and have gome other matters to deal with so I may not be able to respond for a while. Please be patient. Thank you.

[@floramerano3345]

What is the cube root of 20? 2 is my first digit of the quotient. I subtracted 8 from 20 and I get 12. I then bring down the three zeroes now I have 12,000. I multiplied 300 to 2^2 and I get 1200. 1200 x 1 = 1200 which is less than 12000. 1200 x 10 = 12000. How do I deal with that?

[@CrystalClearMaths]

I just found this, your second comment, Flora. Obviously, dividing 1200 into 12000 gives 10, and that is too large a figure for the first decimal place in your answer. If we try 9 (this is just an estimate, remember), then we get the following in the difference column: 30x2x29 = 1740 1740 + 9² = 1821 1821x9 = 16389 Now, 16389 is too large, because we are trying to find a difference below 12,000. This means that we must revise our estimate for the first decimal place to 8 ... and calculate the difference for that! 30x2x28 = 1680 1680 + 8² = 1744 1744x8 = 13952 and this is STILL too large (larger than 12,000). So, we try 7 in the first decimal place: 30x2x27 = 1620 1620 + 7² = 1669 1669x7 = 11683 FINALLY, we have a figure less than the difference of 12,000 and can proceed. So, the cube root of 20 begins with 2.7 As you can see, this is not a simple process and it can require a lot of 'trial and error.' I hope my reply has helped you. Kind regards, Graeme

[@floramerano3345]

Where did you get the "30x2x29"? I thought it was suppose to be "300 x estimate²" which is "300 x 2^2" in this case. And that would equal 1200.

[@floramerano3345]

Also can I also apply this same technique for 4th, 5th, 6th... nth root?

[@CrystalClearMaths]

The number that you find in the estimate column (300 x estimate²) is simply used to estimate what the next decimal digit in your answer will be. By dividing that into the 12,000 our estimate was 'about 10.' Once we had decided to start there, we then test it out in the difference column to see how close the difference REALLY is to the 12,000 figure. This is where the 30x2x29 and 30x2x28 and 30x2x27 calculations take place. We are trying to find a digit that will give us a difference less than 12,000 ... so that we can move to the next step. Sometimes, our first estimate is the one we will use. Sometimes, just as we found here, our first estimate (of 10) had to be revised down until we reached 7. It is frustrating, but that is the nature of this method.

[@CrystalClearMaths]

There are similar techniques for finding other roots. They get increasingly more difficult, of course! They were very useful in the days before mechanical and electronic calculators. Once someone had made the effort to find such a root, they kept a record of it and built up tables they could refer to, rather than have to repeat the process next time they wanted that number. Nowadays, of course, we can calculate these results in seconds using a calculator. Question: How did people design calculators so that they could calculate square and cube roots? That is an interesting thing to consider.

[@lilamjazeefa9466]

Can you show us how to do incomplete polylogarithms by hand? Thanks.

[@CrystalClearMaths]

I must confess, Lilam, that I had/have no plans to create any videos about polylogarithms. My plan was to produce videos to help high school students (and, perhaps, first year university students) gain a better grasp of their mathematics. In time, I do plan on creating videos about reductions of polylogarithms such as gamma functions and, possibly, Bernoulli polynomials, but have no plans to attack incomplete polylogarithms in any way. I am sorry. Thank you for making your enquiry, however.

[@johnwheeler2964]

This is actually a comment on another of your videos on how to obtain approximations to cube roots which I'm having trouble finding my way back to. In it you show three ways to obtain a more accurate approximation to the cube root of a number N if you already have a fairly accurate estemate, a. You call this more accurate estimate b and call the difference epsilon, so that b = a + epsilon. You then show by three arguments that epsilon = (N-a^3)/3a^2. You remark that this becomes a real pain because you must calculate a^3. But you could instead write b = (2/3)a + (1/3)*(N/a^2), which requires only a itself and dividing N by a^2. Moreover, if a < N^(1/3), then N/a^2 > N^(1/3) so the weighted average is closer to the correct answer than at least one of these. Also, if we use epsilon instead to express the fractional error of a from N^(1/3), that is a = N^(1/3)(1+epsilon), then it is easy to show that $b = N^(1/3)(1 + epsilon^2 + O(epsilon^3) ) where O() is the usual "Oh" notation. Thus, the weights 2/3 and 1/3 are optimal in the sense that they eliminate the correction of order epsilon. The same idea works to give b =(1/2) ( a + N/a) = N^(1/2)(1 + O(epsilon^2) for the square root, b = [(3/4)a + (1/4)(N/a^3) ](1 + O(epsilon^2) for the fourth root and so on in the obvious manner. All of these require only the result 1/(1-x) = 1 + x + x^2 + x^3 + ... which is easily motivated by showing that 1+x+x^2 + x^3 + ... = 1/(1-x) applied to (1+epsilon)^n = 1 + n epsilon + (n(n-1)/2) epsilon^2 + O(epsilon^3) It seems to me that this is also an easier way to get to the results for all of the integer roots. As one proceeds to roots above cube roots, the method becomes less practical because the coefficient of epsilon^3 increases, so that the initial guess must be more and more accurate to avoid multiple iterations, and as a result the required integers become more and more unwieldy. John C Wheeler

[@37rainman]

Buddy, you cannot do any better than just learning the long division method, and doing it to its optimum, which he does not do in any of his vids. (No, with all due respect, this guy has not really taught himself the long division method.) Notice in his vids he never does it to an accuracy of more than 3 figures. He always says, or implies that the method becomes too cumbersome to go out to, say, 5, 7 or 10 figures. That, causes a problem, because people accept that, and stop, and never teach themselves the pr5oper method. If you teach yourself to do it properly, you can find roots rounded correctly to 7 figures in 8 minutes. (And you can do that all, w/o erasing, or writing small, on a 4in x 6in piece of paper). To 10 figures in 18 minutes. To 5 figures in less than 3 min. I can even do it to 16 figures on one side of a standard sheet of paper in about 40 min. Sound impossible? Try that with his, or your method. Can you do that with the method you describe here? Btw, nowhere in the LD method do you ever need to cube a, except in the very first (single digit) iteration, which can be done by anyone in their head. So we are not getting the "painful" part. (Actually, in each iteration you need to SQUARE a, but as you teach yourself the method, you will find that that process can be vastly shortened, and eventually, in later iterations, completely eliminated. There are 3 or 4 things, which when you realize them, you will be able to find cube roots to the accuracies mentioned above, and in the time mentioned above

[@CrystalClearMaths]

Thank you, John, for your comment/observation. I am sorry that I have only just found it (You Tube's notifications don't always work). I hope other will read and study your suggestion. When I return to investing time in You Tube again (after a 3-4 year hiatus), I will devote some serious time to it as well. At the moment, I have a 'bit on my plate.' Thank you and kind regards to you!

[@esakkisivabalakrishna2640]

I appreciate the historical reason behind this method. But this method might be needed for competitive examinations especially in India were Calculators are not allowed.

[@CrystalClearMaths]

I agree, Esakkisiva. It is helpful to know how mathematicians learned to calculate these values before the days of mechanical and electronic calculators. If we are asked the same questions in a non-calculator situation (such as the exams that you mention), we must revert to using the same methods that they used. One gains a deeper insight into the mathematics involved, and a deeper appreciation of our forebears and the work that they did. Thank you very much for your comment/feedback.

[@marimuthuselvaraj2646]

Awesome video.

[@CrystalClearMaths]

Thank you, Marimutha. I am glad that you liked it. Kind regards to you.

[@socrates4730]

Sir, i m not being able to find the cube root of 2.9 by your method. Please help.

[@CrystalClearMaths]

Your first estimate will be 1 (the whole number). Since 1^3 = 1, we subtract 1 from the 2 to get 1. Now bring down the 900 from the first three decimal positions. Our new target is 1900. Please let me know if that gets you under way, Prashasta.

[@socrates4730]

Sir, after taking 1900, I find the estimate call which is equal to 1^2 × 300=300. Now nearest estimate will be 6 as 300×6= 1800< 1900. After that I find the difference, [30×a×( a+b) + b^2]b. Here its equal to 3096>1900. Where I m going wrong, sir??

[@CrystalClearMaths]

You are not going wrong, Prashasta. The '6' that you obtained by dividing 300 into 1900 is only an estimate. It is by calculating the difference that we judge if it is a good estimate or a bit too high. In this case, you would then try 5 or 4 as estimates. If you try 5, you will obtain a 'difference' of (30*15 + 25)*5 = 2375 which is STILL greater than 1900. If you try 4, you will obtain a 'difference' of (30*14 + 16)*4 = 1744 which is less than 1900, so your calculation of the cube root has now reached 1.4 and you move to the next digit. Unfortunately, the initial estimates are not always perfect and there is some 'trial and error' involved.

[@socrates4730]

Thanks sir. I got it right this time. The video was very helpful.

[@CrystalClearMaths]

Wonderful! I am glad that I was able to help. Thank you for letting me know. Best wishes for your studies, Prashasta.

[@jerriecoline3679]

This is hard dude

[@CrystalClearMaths]

It certainly is, Jerrie! It gets 'easier' with practice, but it always remains a difficult procedure. We need to respect earlier mathematicians who had to do this by hand before the days of mechanical and electrical calculators. They certainly have my respect! Kind regards to you.

[@jerriecoline3679]

Crystal Clear Maths yeah for me it’s hard so I keep practicing btw thank you for this❤️

[@CrystalClearMaths]

@@jerriecoline3679 You are welcome, Jerrie. I wish I could make this topic easier for you ... but for some topics this is about as 'crystal clear' as one can get :P I hope your practice bears fruit and that it steadily makes more sense and becomes easier for you. Best wishes from Australia!

[@jerriecoline3679]

Crystal Clear Maths thank you💛

[@CrystalClearMaths]

@@jerriecoline3679 You are welcome, friend :-)

[@rezamiau]

Great! thank you.

[@CrystalClearMaths]

You are welcome, Reza.

[@AyushTrivedi-cx2mh]

An amazing explanation given, sir, thank you so much, the way you explained it made it feel like I was enjoying myself and your step by step explanation was very good, The Newton Raphson and Herons method could be used though, however would require the understanding of some concepts Edit - Sir, also how is 8 80?

[@CrystalClearMaths]

Thank you for your kind comment, HGL. You are quite correct about the Newton-Raphson method (for finding any root) and Heron's method (for finding square roots). In fact, the theory behind the procedure that I demonstrate in this video can be derived from the Newton-Raphson method. The video was already quite lengthy, so I did not include these references. I appreciate your taking the time to leave your valuable comment. Very kind regards to you from Australia! PS I am not sure what you mean by 8 80.

[@AyushTrivedi-cx2mh]

I didnt exactly get the part where you actually multiplied it with 300 and not 3, You told, it was a clever way of using the term but arent all of them mixed in the difference.

[@CrystalClearMaths]

@@AyushTrivedi-cx2mh I am sorry for the delay in my replying, Ayush (HGL). I have been very preoccupied with other matters for a long time. I do not have time to devote to KZbin at the moment. Hopefully, that will change during the next few months, and I will be able to reply to your query with some detail. Sorry that I cannot answer your question for the moment. Kind regards to you!

[@archimedesmaid3602]

@@AyushTrivedi-cx2mh This is why he multiplies it by 300 rather than 3 Suppose you are doing this and your first digit of the root is 4. So now you are working on 4b. (You are now trying to find b, your next digit of the root). Your 4 is really not 4, it has actually become 40. So 3 x 40^2 = 4800. 300(4^2) is also 4800 He has transferred the 10^2 to the 3 and made it 300 (4^2) instead of 3 x 40^2. Both ways give the same result -- this is just the method he chose, which is also my method

[@Math-Skill_Venky]

Super video sir.

[@CrystalClearMaths]

Thank you, Math-Skill. You are very kind.

[@haezelgrailgarsi6107]

hello Sir, how about cube root of 0.04,

[@CrystalClearMaths]

Hello, Haezel. I am sorry that I cannot devote time to providing you with a detailed response. You should start by writing the number as 0.040 000 000 and proceed as I have shown. I hope that helps get you under way! Kind regards from Australia.

[@landoc05]

Back in my day you were not allowed calculators or even watches. You had to know this by heart, and all the trigo equivalences, all the integrals, all the periodic table with valences and orbitals, etc.A typical three-hour exam was a simple question of the type: "find the volume of the intersection of the cone described by the points a, c, e, f, g and the sphere described by the points b, c, d, h." And if you failed they put your name on the front door of the faculty with the red note: FAILED.

[@CrystalClearMaths]

Yes. I remember those days. It surprises me that students are encouraged/forced to bring pages of notes to many of their mathematics examinations in Australia. It does not encourage mastery (simply functionality). Thank you for your input and observation, landoc05.

[@anitasingh5089]

Sir u said the above method is "like" long division method . so my question is what is the "actual" long division method. for cube roots

[@CrystalClearMaths]

Hello Anita, What I shared is one form of the long division method for extracting cube roots of numbers. There are variations on this theme, but that are all rather similar. The reason that I said it is like the "long division method" was because, in structure, it resembles the way that we divide numbers using the method that we call "long division." There was nothing more mysterious than that. I hope my explanation helps you and does not disappoint you too much. Sadly, I am not aware of any particularly easy method for extracting cube roots. It is not an easy process, especially if you are trying to produce accurate results. Warm regards, Graeme

[@SpiffyCheese2]

I don't understand what I'm doing wrong!!! I'm trying to find the cuberoot of 52, and at some point I get 2700/25000 ≈ 9 witch would make the cube root approximately 3.9 but the cuberoot of 52 is approximately 3.7

[@CrystalClearMaths]

Unfortunately, I would have to see all your working, TMN. If you try the process again a few hours later, you may uncover some simple error. Kind regards. Graeme

[@SpiffyCheese2]

I've tried the problem about 5 times now and always end up with the same result. I will try to explain what I am doing in the calculation, so first the largest cube that goes into 52 is 3, so I write 3 at the top of the radical, then I put 27(3^3) under the 52, and subtract to get 25, then I add the 3 zero's to get 25,000. Now in my estimate column, I take my estimate(3) and square it to get 9, then I multiply 9 by 300 to get 2700, then I divide 2,700 into 25,000 and get 9 when it should be 7 according to my calculator.

[@CrystalClearMaths]

Now I understand, TMN. Although I spoke of checking one's estimates each time, I realise that I did not make it clear that your first estimate will not always be the best one. Once you have estimated the first decimal to have been a 9, you MUST check that estimate using the difference column on the right. You may follow my explanation from 5:56 in the video. This check is performed by dividing your known digits(s) and your current estimate into parts 'a' and 'b' ... and then calculating 30a x (the entire digit sequence), then adding b², and then multiplying the result by b. In the case that you describe, you have 3.9 at the top (where 9 is your current estimate). The 3 is the 'a' part and the 9 is the 'b' part. We calculate 30a x (the digit sequence), i.e. 30x3 x 39 = 3510, then add b² (81) to get 3591, and then multiply by b (9) to obtain 32 319. We immediately notice that this is substantially larger than 25 000. This means that we now try 8 as our estimate for the first decimal. Starting with 3.8, we calculate the difference as follows: 30x3 x 38 = 3420, adding 8² (64) gives 3484, and multiplying by 8 yields 27 872. Unfortunately, this is still higher than 25 000! Therefore, we need to try 7 as our estimate for that first decimal place. Starting with 3.7, we calculate the difference as follows: 30x3 x 37 = 3330, adding 7² (49) gives 3379, and multiplying by 7 yields 23 653. This IS less than 25 000, so we now know that the cube root of 52 starts with 3.7! We write 23 653 under the 25 000 and subtract, and start the process all over again in order to discover the next digit in the root. As you can see, calculating cube roots by hand can be exceedingly tedious. Fortunately, you will not often have to adjust your estimates in the way that we have had to here. Even so, the method is difficult, and you can see why some earlier mathematicians earned good money by calculating tables of roots and selling them to other mathematicians and engineers and the like before the advent of powerful calculating machines. You can also understand the incredible drive to create calculating machines ... such as the abacus, Napier's Bones, Genaille's Rods, the Pascaline, and many others. I hope this all helps. Best wishes, Graeme

[@CrystalClearMaths]

Something extra for you, TMN. If you watch my explanation at the end of the video (from 14:44), you will learn why your first estimate was not especially accurate. This was because the 3 was a smaller digit and the 9 was a large digit. If you watch this (algebraic) explanation, I hope you will find it informative.

[@SpiffyCheese2]

Thank you!!!!! :D Learning and understanding how to perform such calculations are very useful for helping me develop hardware and software for my computers. Once I watch your video on how to calculate the n'th root of a number, I can now progress in my hardware development. I'm 14 years old and has loved math more than anything in the world since the age of 3. I am currently trying to go to college next year and take calculus 2. I learned most of my mathematics from khan academy, but I still have questions that I need to be answered and you have helped me answer one.

[@jaishree1886]

Can we also find out the fourth root of any number without using logtables

[@CrystalClearMaths]

Absolutely, Jai. The process is quite daunting, however. I have another video in which I generalise this and explain how to calculate the nth root of a number (in principle). You might like to watch it ~ kzbin.info/www/bejne/bHLVo6uboL6FaLc. Best wishes to you.

[@CrystalClearMaths]

Very true, tampicokid. You show good insight! There is one restriction or problem with taking the square root of the square root, however. The principle can be illustrated like this: If you require the fourth root to an accuracy of, say four significant figures, then you must calculate the first square root to eight significant figures. Despite this, your suggestion may still provide the simplest route to calculating the fourth root anyway. Thank you for your observation and for taking the time to comment/share. Best wishes, Graeme

[@CrystalClearMaths]

Thank you, tampicokid. I will try it when I have time. I will admit that I based it on the idea that the square of a four figure number is (roughly) an eight figure number. Thank you for your input. Others can experiment as well.

[@CrystalClearMaths]

Thank you, tampicokid. I hope other viewers will explore this, too. I must confess that it is an area that I had not bothered to investigate. I appreciate your input.

[@CrystalClearMaths]

I will certainly examine your post in detail when I have the time, tampicokid. Thank you for taking the time to post such a detailed analysis of the process. At this stage, I have produced no new material on KZbin (or the Internet) for 2.5 years as there has been a lot happening in my 'real life.' It may, realistically, be another 6-12 months before I can devote myself to this material again. In the meantime, I hope that other viewers will pursue and explore the recommendations that you have made. I appreciate your contributions. Thank you. Graeme

[@holyshit922]

I calculate cube root in this way Let a be actual estimate Let b be last digit of next estimate (10a+b)^3=1000a^3+300a^2b+30ab^2+b^3 (10a+b)^3-1000a^3=300a^2b+30ab^2+b^3 (10a+b)^3-1000a^3=(300a^2+30ab+b^2)b (10a+b)^3-1000a^3=((300a^2+b^2)+30ab)b In first step I choose a in that way the difference be smallest nonnegative number in next steps i choose b in that way the difference be smallest nonnegative number but maybe your way is a little bit better from computational point of view

[@CrystalClearMaths]

Thank you for your contribution, Jacek. KZbin did not notify me about your comment and I have only just found it. I appreciate your taking the time to share your insight/approach relating to the difference between cubes. Kind regards from Australia.

[@saritarawat1681]

how can i calculate the cube root of "15" by using this method. every time i start doing it i get stuck in between the steps. plzzzzz upload a video showing the calculation of cube root of "15". Plzzzzzzzzzzzzzzzzzz

[@CrystalClearMaths]

Sadly, Sarita, I will not be able to produce videos for a while yet. There have been a lot of things happening in our lives here and, at the moment, it is just not possible. Hopefully, your friends or teachers may be able to help. Otherwise, you may simply have to watch and pause the video a few times and try to work it out for yourself (actually, you will remember it far better if you manage to do this). I am sorry that I cannot assist you at this stage. Best wishes for your studies.

[@37rainman]

He shows the method on his vid. For 15 it is exactly the same. Look closely , In each iteration, the divisor is simply a combo of the current root squared, and multiplied by 300, and the currant root times 30. The first digit in your example is 2, as 3 cubed is 27 -- to big. If you concentrate you can easily conquer this algorithm. Also, it looks daunting to work out the cube root for instance to 10 places, but actually it is not. There are 3 things you can discover which will greatly abbreviate the process. I can do roots rounded correctly to 10 digits in 18 to 20 minutes. To 7 digits in 10 minutes. To 4 digits in 4 minutes. Using nothing but pen and paper. What are those 3 things you can employ to shorten the process? +Crystal Clear, have you discovered them?

[@akshaykumarpandey7131]

hello sir your video was nice but i had a doubt in the second case why you use a=8 for estimate but you use a=80 in the difference?

[@CrystalClearMaths]

Thank you for your question, Akshay. Once we added the 5 as the second part of the estimate, it was then 85 (ignoring the decimal point). Since this was the (a + b) that we used, the a = 80 and the b = 5. I hope this makes sense. The theory presented during the last third of the video may make the reasoning more clear. Best wishes to you.

[@zarinefathima1291]

Thank you very much sir

[@CrystalClearMaths]

You are welcome, Zarine.

[@akashpq]

धन्य हो गुरु जी

[@CrystalClearMaths]

You are welcome, Akash.

[@user-Danger696]

Very nice sirrr

[@CrystalClearMaths]

Thank you, Aditya. Good to hear from you again. Best wishes.

[@user-Danger696]

@@CrystalClearMaths welcome sir keep your blessings on me

[@CrystalClearMaths]

@@user-Danger696 :-)

[@floramerano3345]

What about cube root of 20?

[@CrystalClearMaths]

This method works for the number 20 as well, Flora.

[@pazil888]

Thank you for watching ? I thank YOU for showing ME !

[@CrystalClearMaths]

In that case, we are both blessed, Liza :-)

[@Mann..]

That's boomer......👌🏻👌🏻 Amazing

[@CrystalClearMaths]

:-) You are welcome, Mann. I'm glad that you liked it!

[@royalrahulgaming1495]

how to find cube root of 503

[@CrystalClearMaths]

Hi Mast Pic, you should be fine if you follow the instructions in this video. If you are still having trouble, ask a teacher or explain to me what you have achieved so far. You will remember better if you follow instructions and work it out for yourself.

[@peterbayonet1266]

This method fails when I tried to calculate the cube root of 2. The result should be 1.25992..., but this method yields the second digit as 3...

[@CrystalClearMaths]

You will quickly find, when you test the estimate for the second digit, that 3 will be too large, so you will have to revise your estimate to 2 and proceed accordingly. Unfortunately, Peter, the initial estimate is not always exactly right (it is an estimate only). The real test comes when you calculate the difference. I hope this makes sense. Kind regards to you (and thank you for your question).

[@jaishree1886]

Thank u sir u r awesome

[@CrystalClearMaths]

You are welcome, Jai. Thank you.

[@rishabhlakhara9583]

Thank you sir

[@23jeninaangelind35]

Nice sir

[@CrystalClearMaths]

You are welcome, Jenina.

[@venkataramanan8470]

Love from india

[@CrystalClearMaths]

Warm regards from Australia :-)

[@venkataramanan8470]

@@CrystalClearMaths thank u

[@CrystalClearMaths]

@@venkataramanan8470 You are welcome!

[@muskamelogbi4716]

God bless you and protect your own

[@CrystalClearMaths]

Thank you! May God grant you His grace and peace, too!

[@aaishuvlog620]

please u also tell me the rational no.

[@CrystalClearMaths]

Hello Aashiya. I am sorry, but I am not sure what you are asking. Radicals (roots) cannot normally be expressed as a fraction. Their approximate value can be written as a fraction, however. For example, if √623 is approximately 8.54, then we could write this as 854/100 = 427/50 (or 8 27/50). Therefore, this is a rational number with a value close to √623. I am not sure whether this is what you mean.

[@melbymathews4207]

still helps me

[@CrystalClearMaths]

I am glad, Melby. Kind regards to you from Australia.

[@udaypratapsingh3503]

Nice I learn I subscribed

[@CrystalClearMaths]

Thank you very much, Epic. I'm glad that you enjoyed this video and hope you enjoy others on this channel. Unfortunately, because of some health and family matters, it will probably be another 6-8 months before I start producing videos again. When they start appearing again, I hope you will enjoy what I have planned. Best wishes to you.

[@muskamelogbi4716]

Dear my friend may l ask you à question how optimisation calculus could be a part of infinitesimal caculus ? Thanks lot.

[@CrystalClearMaths]

The infinitesimal calculus is the theory and process for finding derivatives (or gradient functions). It is the tool that we use for optimisation when we locate maximum and minimum values of functions (where the gradient function has a value of zero).

[@bonbonpony]

This method gets pretty intractable pretty fast, because the numbers grow by three digits each step. As if long division weren't long enough already... :q And all of that hard & repetitive work just to get _two_ decimal digits of precision? ;o

[@CrystalClearMaths]

I agree, Bon Bon. You can understand the race (by mathematicians in the past) to work out more efficient methods for calculating roots, trigonometric, and logarithmic values, for example. Those who were extremely good calculators invested considerable time in compiling books of tables of values. Some made a good income from the endeavour. We can be thankful that we now have electronic calculators!

[@archimedesmaid3602]

Actually when you get experience with this method, and figure out some further things, you can very rapidly evolve CR to large amounts of digits in little paper space and time. Youve just got to get some experience and figure things out. Sounds unbelievable, but I once demod that I can find the root to 25 digits on one side of a sheet of paper. No calculators, no separate worksheets, no erasing at all. Simple addition, subtraction, multiplication -- no special abilities -- gradeschool maths I hear you saying: "But the last digit in the long division problem will be something like 70 digits long!! Impossible on 1 sheet, impossible on 50 sheets, and besides that, it would take a month!!" Its all on 1 sheet. (Well it did take 2.5 hrs). So what do I do???

[@bonbonpony]

@@archimedesmaid3602 May I see that sheet?

[@archimedesmaid3602]

@@bonbonpony Well.....................(-; Wouldnt that ruin your joy of figuring it out for yourself............ !!?? (-; How many figures have you evolved a CR to so far? If you do this awhile I am thinking some obvious things will suggest themselves I could give you some cryptic hints! For one thing, in each iteration you need to square ever increasing numbers. Think about it! What can you do with this fact? Most major hint. I decide to how many places I want to evolve my root , BEFORE I ever start. (In other words, I am suggesting that if I did it to, say, seven places, that work would look far different than what the work to seven places looks like on a root taken to say, 13 places. Another, and related,, and REALLY major hint! (Fact is, I have likely hinted it away). 🙂 Lets say you are trying to find the root of 345 321.005 781 111 629 000 222 051 963 to 10 figure accuracy. What statement can you make about the majority of those numbers there? The example to 25 digits is in my cellphone photos, but nah, cant send it to you. Wouldnt be right........🙂🙂

[@archimedesmaid3602]

Hilariously, I just came across your reply of over a year ago. If you provide me with an email address, I can email you a picture of that page. It is still in my phone images Sorry about the huge delay 😅

[@piyushchoudhary6414]

Please tell of 0.25

[@CrystalClearMaths]

I do not understand what you mean, Piyush. If you are asking how to calculate the cube root of 0.25, simply prepare the same groups of three digits first (i.e. 0.250 000 000) and proceed from there. The position of the decimal point does not really create any difficulty.

[@aaishuvlog620]

Nice video

[@CrystalClearMaths]

Thank you, Aashiya.

[@sillymesilly]

I would say Babylonian method be easier?

[@CrystalClearMaths]

It very likely would be ... a good iterative technique. Thank you for your astute observation, Silly Me Silly.

[@malazeshafurfur5291]

Not in NMAT sir... we dont use calculator in National Medical Admission Test. 😭😭😭 I wanna cry, Im so dead😭😭

[@CrystalClearMaths]

Oh dear. It sounds as though you have a huge challenge ahead, Zesha. How long do you have to prepare?

[@malazeshafurfur5291]

Crystal Clear Maths our exam is this coming march 26. 😭 I dont have enough time.

[@CrystalClearMaths]

That is very soon, Zesha. I have never seen a NMAT but I looked it up and you must live in the Philippines. I cannot recommend what to study but encourage you to get sufficient exercise, good food and sleep, especially in the days before your tests. Sitting for your tests while tired can decrease your score considerably. Having a 'fresh and clear brain' will be one of your biggest assets during each exam. Studying in intensive bursts can make your study a bit more effective, especially if others can work with you (for example, mentally solving problems while a friend checks your answers ... and vice versa). I do not wish to tell you what to do but those principles may help. All the very best for your studies. Please keep in touch and let me know how you go.

[@jaishree1886]

Now I don't have to go for the log tables for finding cuberoots

[@CrystalClearMaths]

Very true, Jai ... but how time-consuming this method is! In fact, this is why some mathematicians made good money many years ago by constructing tables of such values and selling them ... to save others from all this labour! They were recognised as 'calculators' (that is what people called them).

[@jaishree1886]

@@CrystalClearMaths sir i love maths and i want to learn everything infact if god appears in front of me to fulfill my wishes i would ask for a complete knowledge and skill in calculus

[@CrystalClearMaths]

Jai ... you have an honourable passion. You may find my calculus videos useful, too (see my playlists). I wish you well with your studies. Graeme

[@ronalddump4061]

@@CrystalClearMaths This method of finding cube roots does not need to be as time consuming as you suggest. You simply have not progressed the method to its logical conclusion. When you do that you, like i, will be able to evolve the cube root of any number to 10 digits accuracy in 15 to 20 minutes, on 2/3rds of a sheet of paper. To 7 digits on a 4 x 6 inch card in 8 to 10 min. (And note: In the process there are NO "savant talents" involved. No, say for example, squaring 3,4,5,6 digit numbers in ones head. Just a whole bunch of boring, elementary addition, subtraction, multiplication. Sounds utterly impossible, correct? It is not. It is just that you havent really thought about it, and brought your method to a logical conclusion. And i am truly wondering why.... Its been, what? 5 years?? If you provide me an email address i can send a few examples to you for your inspection. What is the CR of 79 819. 300 912 256 111 696 555 777 rounded correctly at the 10th digit? You can do it in less than 20 minutes, pen/paper, NO calculator involved before or during. But first, you guys need to sorta step out of that box you are stuck in........

[@CrystalClearMaths]

@@ronalddump4061 You may well be correct, Ronald. Please post a video showing your method and insert the link here. It would be valuable for viewers to see alternative methods. I had used this approach to follow closely the binomial expansion (as a principle) and to show the link with calculus as well. Thank you for highlighting that alternative methods exist.

[@darogajee3286]

hey man just tell the generalised method for calculating nth root of any number

[@CrystalClearMaths]

Hi Daroga. I have a separate video where I do just that ~ kzbin.info/www/bejne/bHLVo6uboL6FaLc. You made a good point. Thank you.

[@sri8820]

hey..do u know that n th root method.. please give any link for that video..

[@CrystalClearMaths]

Hello, Sridevi. I only just found your comment (KZbin did not notify me about it), so I am sorry for my late response. You may find this link useful ~ kzbin.info/www/bejne/bHLVo6uboL6FaLc. Kind regards, Graeme

[@aksmath6087]

i have found my own formula to calculate the cube root of any real number by division method, now i am going to find the nth root formula ...

[@aksmath6087]

@@sri8820 i have found my own formula to calculate the cube root of any real number by division method, now i am going to find the nth root formula ...

[@holgerdvachlis6560]

I think I can build my own nuke now (on paper) XD

[@CrystalClearMaths]

Hahahahaha :-)

[@antoniolaurito816]

There's another way to calculate cube root that slightly different from the video

[@CrystalClearMaths]

I am aware that there are other algorithms, Antonio. I used this one as it clearly illustrates the use of calculus or of expanding a binomial expression. The other methods are also based on these two principles. If you are happy to share the method that you know (have seen), I would greatly appreciate it. Thank you.

[@sourav2018]

thanks sir

[@CrystalClearMaths]

You are welcome, Sourav.