Extreme Algebra Question (

If a+b+c=1, a^2+b^2+c^2=2 and a^3+b^3+c^3=3, then what is the value of a^5+b^5+c^5? Isn't it just 5?? Of course, you will have to watch the video to see how to solve this extreme algebra problem! By the way, this is a very common type of algebra problem in math competition exams. Check out the following related videos
binomial expansion formula, expanding (a+b)^2, • How to expand (a+b)^n ...
trinomial expansion formula, expanding (a+b+c)^2, • the easy way to expand...
‪@blackpenredpen‬

[@blackpenredpen]

EXTREME CHALLENGE: Find a, b, and c individually!

[@fayezdeadpool8993]

:/ hold my bear

[@SameerChandra]

Next video hopefully?

[@Davidamp]

They are the solutions of a 3th degree polynomial!

[@ashwinvishwakarma2531]

no

[@maximunnit]

@@Davidamp 3th

[@Andrew..J]

I'm just gonna say 5 and hope the rest of the test is easier.

[@maxwell_abe2894]

Same, I just do. Eny meany myny mo( don’t think anyone knows how to spell it) and what ever it lands on that’s the answer lmao. I do that for 99% of the test

[@Mukawakadoodoo]

What if it’s the only question on the test 🤣🤣🤣

[@Me-vu3yv]

@@Mukawakadoodoo then you are in trouble

[@naturalaviator4156]

Lmao

[@parthibhayat]

I did actually try to solve on my own, I got 5 at first cause I forgot to make the minus into plus after changing the side of some expression. Took me some while to figure out where I got it wrong

[@rob6129]

hes becoming a real professor: occasionally confused by himself

[@blackpenredpen]

R Hahahhahahaahahaha right on!

[@mystudy6827]

Hes becoming a real professor:

[@Pygmygerbil88]

@@mystudy6827 isn he?

[@LORD-kg4sl]

He is already a teacher lmao

[@nickpapagiorgio9872]

LORD hopefully he learned to slow down so people can understand . It’s not a race.

[@mustafakhurramaziz9715]

Me:*looks at whiteboard* Whiteboard: 0:30 Me:*looks outside the window for 0.000001seconds The whiteboard: 40:04

[@XoIoRouge]

It's a shame the joke has to use timestamps. Wish it could've used pictures. 😂

[@deleted-something]

True

[@exzar9684]

in class everytime 🤣🤣🤣

[@wanlitan7406]

0:14 "We know the fourth power is going to be too easy" Uh, excuse me?

[@darionsama6612]

*pauses the video to try to solve it* *notices the 41 minutes length* *plays the video and throws the notebook away*

[@hrutikabadgeri6505]

😂😂😂

[@NovaWarrior77]

I applaud you! Glad I was not the only one who tried to solve it a little.

[@oximas]

i triend to solve it then realease it would take for ever to solve the problem and i am getting nothing afterwards

[@rishabhbharti3580]

🤣🤣🤣🤣🤣🤣🤣🤣

[@ashumishra7020]

Actually he had to talk a lot and stumbled a few times while writing it on board that's why it took him soooo long. It solved it on notebook and took me some 14-15 mins (actually a lot less I think )... But I really leaned some thing new .. the Pascal pyramid

[@LuisMartinez-ih2oc]

Teachers example: 4x-3y=6 y=-3x+15 Classwork: x-3y+3z=-4 2x+3y-z=15 4x-3y-z=19 Exam:

[@ilya9481]

Exam: prove a^n + b^n ≠ c^n (n>2)

[@justaconcrete4789]

@@ilya9481 Hahah that's the last theorem of Fermat's, Incredible brain to solve it

[@adityashankar5723]

@@ilya9481 I actually started solving that.

[@GodOnVayne]

(5/3, 44/9, 3) (x,y,z)

[@bencroshaw7293]

@@GodOnVayne . Did you use matrix multiplication to work the equation out?

[@blazingfire7517]

“Try this first” Me seeing that this is a 40 minute video: “no, I don’t think I will”

[@andyb6177]

"Try it for yourself" *opens MatLab* *gets answer with less then 10 lines of code*

[@beechass4451]

B-b-but that's not the point man :(

[@78anurag]

Code pls

[@darrylchandra554]

But TBH it ain't hard, it is just complicated

[@AryanSingh-yb2my]

How did you write the code, could you share yours, ive been trying to make it but I just cant:( send yours pls

[@Nickesponja]

If I ever become a college professor, I'm going to put this as a common exercise and tell my students it's trivial

[@General12th]

"The proof is an exercise for the reader."

[@jahad6335]

But... it is actually trivial

[@General12th]

@@jahad6335 If the derivation takes forty minutes, it's not trivial. You're just trying to make yourself look smart. Cut it out.

[@Nickesponja]

@@General12th One time a professor told me that proofs start being not trivial when they take more than 3 pages xD

[@alessandrofenu4325]

@@General12th well..the fact is that this is actually easy. It's not me bragging around. Yeah okay that's long but it's literally a standard exercise with no particular idea behind (still big up to blackpenredpen for doing that video).

[@nicholaswongso2003]

bprp: "find a, b, and c individually!" me: *no*

[@505steel]

I love how he is always so excited about math! Such an inspiration

[@hatasesasd9631]

15:55 * Erases K *... No, It's C to the k power * Writes a new K *

[@ngochoang4639]

Lol

[@BlankTH]

lmao ye and the way he goes "whew" afterwards like thank goodness I noticed. omg my sides

[@arah8998]

He fixed it in minute 20 (20:00)

[@Nio10895]

Thats so me

[@albusdumbledore2406]

That ruined my ocd

[@blackpenredpen]

Mistake at 3:06 . a^2*c^3 should be a^3*c^2 but the rest is okay.

[@vitakyo982]

I was going to comment , but ...

[@leadeer4213]

15:50 u wrote k no c in base

[@omarmachuca6405]

blackpenredpen How is this written 2 days ago?

[@leadeer4213]

@@omarmachuca6405 That video wasn't public

[@sarthakgarg5701]

at 19:00 its beggar method or multinomial theorem . answer is i guess n+r-1Cr-1 where n are the total things which are to be distributed among r people. please correct if im wrong. My teachers at school taught me this a year ago

[@jonathandavis8014]

This reminds me of when I was proving the Marion Walter Theorem via Coordinate Geometry and I had to solve an 8 variable equation and thus produce and inequality afterwards. Awesome content dude!

[@Mqxwell]

Absolutely loved this. Would love to see more.

[@pleasepassthemustard4401]

That one kid in math class who does this even when the teacher says you don't have to show your work

[@blackpenredpen]

Please pass the Mustard hahahaha

[@HappyVLR]

Xd

[@toopytoopy8547]

Hi! I found an answer to your problem without heavy calculus. Let's consider the polynom P(X)=(X-a)(X-b)(X-c) By developping P, we obtain that P(X)= X^3 - (a+b+c)X² + (ab+ac+bc) X - abc We have : (a+b+c)²=(a²+b²+c²)+2(ab+ac+bc).So, ab+bc+ac=-0.5 Let d= abc . Therfore, P(X)= X^3 - X² -0.5 X -d (*) As a,b and c are the roots of P, then : P(a) = P(b)= P(c) =0 Then , using (*) : 0=P(a)+P(b)+P(c)=(a^3 - a² -0.5 a -d)+(b^3 - b² -0.5 b -d)+(c^3 - c² -0.5 c -d) =(a^3+b^3+c^3)-(a²+b²+c²)-0.5(a+b+c) - 3d = 3 - 2 -0.5*1 -3 d Then d=1/6 Now, let's calculate a*P(a)+b*P(b)+c*P(c) We have then : 0= (a^4+b^4+c^4)-(a^3+b^3+c^3)-0.5(a²+b²+c²) - d (a+b+c) So, a^4+b^4+c^4 = 3+0.5*2+(1/6)*1= 4+1/6= 25/6 Finally, let's calculate a²*P(a)+b²*P(b)+c²*P(c) We have then : 0= (a^5+b^5+c^5)-(a^4+b^4+c^4)-0.5(a^3+b^3+c^3) - d (a²+b²+c²) So, a^5+b^5+c^5 = 25/6 + 0.5*3 +(1/6)*2 = 6 ;) Using this recursive method, we can calculate a^6+b^6+c^6 then a^7+b^7+c^7, etc I hope I did any calculus mistake :)

[@boong0006]

Thank you!

[@boong0006]

Since P(X) is a trinomial, I wonder if you can apply this method to a quadnomial that involves a,b,c and d?

[@dacres2002]

Thats the best comment ever!!

[@dacres2002]

In the last line you got a mistake. It is not 16/3 but 6. This is an amazing method, because you can find other sum of powers

[@toopytoopy8547]

@@boong0006 Yes. You can apply this method to any degree. In fact, if the problem was to find (a_1)^(n+1)+ (a_2)^(n+1)+ .....+(a_n)^(n+1) such that you know the values of the sums s_n := (a_1)^k+(a_2)^k+...+(a_n)^k for 1 you compute p_{n-1} s_2 + p_{n-1} s_1 + 2 p_{n-2} = 0 >> you compute p_{n-2} s_3 + p_{n-1} s_2 + 2 p_{n-2} s_3 + 3 p_{n-3} =0 >> you compute p_{n-3} . . . s_n + p_{n-1} s_{n-1} + 2 p_{n-2} s_{n-2} + .... + (n-1) p_1 s_1 + n p_n = 0 >> you compute p_n Now, you want to find s_{n+1} : =(a_1)^(n+1) + ...+ (a_n)^(n+1) Let's calculate a_1 * P(a_1)+...+a_n*P(a_n) we have then 0 = s_{n+1} + p_{n-1} s_n + p_{n-2} s_{n-1} + ...+ p_1 s_2 + p_0 s1 Thus, we deduce the value of s_{n+1} To compute s_{n+k} (where k>1) recursively we calculate (a_1)^k * P(a_1) +...+ (a_n)^k * P(a_n). I hope that was clear :) It's a pity that KZbin doesn't allow mathematical writing.

[@jrajesh11]

Brilliant working out impromptu on board. Really good!!

[@robertokang7557]

Amigo, amo tus videos, aprendo mucho. Aunque no se mucho ingles,pero tus explicaciones son mejor que muchos catedráticos que explican matemáticas.

[@JoshuaHillerup]

I just posted this on my Facebook as If 🍋 + 🦆 + 🌳 = 1, and 🍋^2 + 🦆^2 + 🌳^2 = 2, and 🍋^3 + 🦆^3 + 🌳^3 = 3, what's 🍋^5 + 🦆^5 + 🌳^5? I'm curious what answers I'll get.

[@blackpenredpen]

Joshua Hillerup hahahhahahahha

[@richardbembridge1822]

Brilliant.

[@1remphase1]

Winner 🙈

[@fourier07able]

For sure they change your symbols by a, b, c, or x, y, z, etc.

[@VENOM-tx6gp]

Wait you MUST NOT forget to write 98% fails to solve

[@vibhavaggarwal237]

blackpenredpen did it in 40 minutes. How about I calculate both a⁴+b⁴+c⁴ and a^5+b^5+c^5 in less than 3 minutes? Ready for some fun? Okay, stay with me. Let us assume that a,b and c are zeroes of some cubic polynomial f(x). So we have f(x) = (x-a)(x-b)(x-c) = x³ - (a+b+c)x² + (ab+bc+ca)x - abc Now our task is to simply find those coefficients. a+b+c = 1 2(ab+bc+ca) = (a+b+c)² - a² - b² - c² = 1² - 2 = -1 Hence, ab + bc + ca = -1/2 Also we have the following identity: a³+b³+c³ - 3abc = (a+b+c)(a²+b²+c²-ab-bc-ca) Thankfully, we know other quantities so it gives us abc = 1/6 Hence our polynomial is f(x) = x³ - x² - (1/2)x - 1/6 You still here..?? Good! We are almost there. Check your watches. Since 'a' is a zero of f(x), a³ - a² - (1/2)a - 1/6 = 0 After multiplying both sides of this equation by 'a' and a little bit of rearranging, we get: a⁴ = a³ + (1/2)a² + (1/6)a We will get similar equations for b and c. After adding them, a⁴+b⁴+c⁴ = (a³+b³+c³) + (1/2)(a²+b²+c²) + (1/6)(a+b+c) = 3 + 2/2 + 1/6 = 25/6 Similarly, a^5 = a⁴ + (1/2)a³ + (1/6)a² Adding again, a^5 + b^5 + c^5 = 25/6 + 3/2 + 2/6 = 6 Voila! I appreciate this video nonetheless.

[@peterderias5323]

Yea I was taught this way in high school and it blew my mind for a solid week because I did an entire excersise without it and died of the monotony

[@vibhavaggarwal237]

@@peterderias5323 That's interesting :)

[@gekkouga5147]

I didn't get it😥

[@vibhavaggarwal237]

@@gekkouga5147 Have you studied polynomials? If not, I would suggest to study that first. If yes, tell me which part is troubling you.

[@SonGoku-rx1fo]

I got it what you said and I believe only top level of people who mastered algebra can solve it and understood it. At least write the values of a+b+c and other from video to make it more easy to understand. Well done btw but all I hope that you haven't used Google for this question. 🤔 I solved it too but I forgot to assume f(x) = x"3 - x"2-........ That you did. I did (a+b+c) (a^3+ b^3 + c^3) which gives a^4.....+c^4 = 3 ;(. Later I thought to do some assumptions like yours and tried but again I forgot to add trinomial formula of abc ;( and i gave up. Btw I did this much calculation in my mind in 3-4 min so I failed 😅.

[@t.k.-s.4212]

Someone who is dedicated to Algebra as much as I am! Thank you for this video. Just discovered your channel!

[@raphielohnef4678]

2:00 You can solve the question by taking all the products of the original formulas that yield a^5, b^5 and c^5 as summands (those are 5 products). Interestingly, you get 5 different types of summands as well (a^5, a^4 * b, a^3 * b^2, a^3 * b * c, a^2 * b^2 * c). Thus, you get a set of 5 linear equations and 5 variables (the number of times the types of summands appear). Solving this so you only get the first type of summand also gives you the solution 6.

[@micronalpha]

Congrats on showing the non-edited video. We've all been there, batteling hard algebra. I enjoyed seeing this video. ☺

[@blackpenredpen]

Mauro Maia thank you!!!!

[@anshumanagrawal346]

@@blackpenredpen I found a way to solve this problem which is very easy and short

[@dux2508]

Anshuman Agrawal Please share

[@MrKrabs-xf2tr]

@@dux2508 Simply use Newton's Identities and consider a,b, and c as roots of a cubic

[@anshumanagrawal346]

@@MrKrabs-xf2tr Yes, exactly

[@sinzuii2061]

Skipping from 0:00 to 28:00 Is like my teacher after i took a restroom.

[@WilliMel]

@@codeguy21 dafaq

[@ifteharulhaque2600]

@@codeguy21 wut?

[@fdnt7_]

@@codeguy21 TF?

[@molozu6370]

damn bro pass me a restroom too

[@jiauddin007]

that means you go to the restroom for 28 minutes.

[@iankr]

Got there in the end. This guy is so endearing, you can't help but love it.

[@iliasghannam9252]

Thank you so much it was incredible to watch. I love the way he discovers the small mistake due to a la lack of attention he just stops speaking for a while. I thought the video will be long but actually it wasnt. Thanks

[@georgemissailidis1504]

And what sucks is that if you make *one* mistake, everything to follow is all wrong.

[@NavsangeetSingh]

Funnily enough, he made one mistake in the beginning around 3:08 mark. But somehow ended up getting the right terms in the next step just because of the ongoing pattern.

[@stickman32032]

Proud to admit I watched the video all the way through instead of doing my homework

[@d4670]

One if my favourite videos, I love how long it takes to get to the answer

[@nol2521]

Man, what a perfect example of what it feels like when the teacher asks you to show your work, when you got your answer based off your intuition

[@Debg91]

Awesome! Your exercises are getting more and more interesting!

[@Arbmosal]

The Fundamental Theorem of Symmetric Polynomials seems to make this very easy and also pretty straight forward. I have not double checked it yet, but my rough solution seems ok.

[@nikhilnagaria2672]

I remember watching it about 2 years back. Man, how much have I grown, how much has changed from then

[@alex3256]

Didn't think there was any way I would watch this whole video for what looks like a simple question. But this guys energy got me hooked xD

[@BrainGainzOfficial]

Patience is key! Love these kind of videos man, keep it up!

[@stevehof]

I love this guy! He makes math more entertaining than a lot of Hollywood movies I've seen lately!

[@animeshdebnath594]

What a great question! Thanks a lot for sharing it

[@JonathonV]

Some people were wondering how you could find the value of a⁴+b⁴+c⁴ without tools such as Wolfram Alpha. A few people were actually pretty snippy about it, but it can be done, and if you have the right approach you can do it without too much trouble. Here’s what I did: I started by expanding (a²+b²+c²)² and substituting a²+b²+c² = 2. Some minor rearrangements left me with what I'll call equation A: a⁴+b⁴+c⁴ = 4-2(a²b²+b²c²+a²c²) Next, I expanded (ab+ac+bc)². From the expansion of (a+b+c)² we know that (ab+ac+bc)² = (-1/2)² = 1/4, so we get that 1/4 = a²b² + b²c² + a²c² + 2abc(a+b+c). Substituting a+b+c=1 and the fact from the video that abc=1/6, we get that a²b²+b²c²+a²c²=-1/12 (which happens to also be the sum of all the natural numbers, but I digress). Swap that result back into Equation A above, and a⁴+b⁴+c⁴=4-2(-1/12)=25/6.

[@gileadedetogni9054]

How interesting, it's approximately 4

[@auroremelie]

You should invest into a small microphone that you can attach to your clothe.

[@naman334]

na its iconic at this point

[@luismariabiaggioni8514]

It is a grenade. It si not a microphone. Ja ja ja

[@rahuljadhav7156]

@@luismariabiaggioni8514 lmao

[@ishworshrestha3559]

Ok

[@agnibeshbasu3089]

NAH! then how he will show off his black pen red pen trick?

[@virakboththan3550]

This is so amazing I’m so surprised

[@user-zm7nv9in5r]

We can get abc more easily by using (a+b+c)(a^2+b^2+c^2-ab-bc-ca)=a^3+b^3+c^3-3abc

[@somenamelastnaammee52]

1)Subtract eqn 3 and 1. 2) Equate the terms with eqn2. 3)U get 3 solutions for x , those are for x,y,z 4) profit

[@veetee355]

What the hell is this.

[@jasonstarrising]

vtt355 It’s algebra 2 but like the accelerated version.

[@gregoriousmaths266]

Craziness

[@AZ-ps1si]

Our Worst dream.

[@user-iz5qq9ht1m]

Abdulaziz Alquhaibi عبدالعزيز هذا شنو 😂

[@nameless5678]

A nightmare

[@Andy-ju8bb]

Just watched my 3rd back-to-back blackpenredpen video. My brain now needs a vacation on a tropical island.

[@blackpenredpen]

Andy awww : ))))

[@youniquelol5850]

i hop on youtube to get a break from my math homework, and i see this on my recommended! at first i clicked on as a joke, but even half way through i had learnt a lot, and i was actually invested in learning about this. great video!!

[@chemadepot2392]

this is so fun. I love watching your channel

[@isabahk1132]

*If this was my exam, it would just be worth like 6 marks, nothing else*

[@duggydo]

38:00 implies complex solutions! 😄

[@blackpenredpen]

duggydo pretty much!

[@typo691]

What makes you say so? Experience or memes?

[@blackpenredpen]

Typo ???

[@reetasingh1679]

@@typo691 Its obvious... The left side of the equation has squared terms and the right side is negative. This is only possible if the left side had terms which had the square root of -1.

[@keescanalfp5143]

@duggydo, From the first minute we thought of a kind of symmetrical couple, triple, in the complex plane.

[@ayceeren]

I love how u keep uploading dis kind of videos!! u r amazin.

[@bukstv6340]

Thank you sir, I learned a new way of expanding trinomials

[@davidnobre171]

Around minute 32 when you look at the board and said something's wrong shit. That legit is me whenever I get a negative value for time after a page of calculations and now have to go to the beginning look where I lost the sign

[@milindbordia]

David Nobre don’t worry, time is relative

[@x15cyberrush9]

hahaha

[@crosisbh1451]

That's when I inconspicuously add/drop a negative somewhere in my work to just make it work out.

[@keescanalfp5143]

@David Nobre, Algebra is outside time domain, algebra is beyond time, algebra is timeless. Examinations aren't. Algebra books aren't. Olympiad sessions aren't. Even blackpenredpen videos aren't. Regrettable? Well, uhh, rr. **

[@ffggddss]

There's a theorem about mistakes in long calculations: "The length of time it takes to find your mistake is inversely proportional to the number of possible answers." And the corollary to that, is: "Finding the sign of a result takes the longest possible time." Fred

[@user-qd4kt7ze3o]

Watched the whole video + spent 20 minutes trying to solve it myself(didn't succeed 😢). Unbelievably entertaining.

[@chemiflask7692]

I love math, good job. I cracked up with this problem. I'll try it. Love your channel!!!

[@adaakdeniz6033]

i tried to got all the points and i had so much fun ! thank you so much for this video 😌🙏🏽🤩❤️

[@goup867]

Well after you find a ab+bc+ac and abc you can pretty much solve for nth power constructing third degree polynomial with a,b,c as roots (x^3-x^2-1/2x-1/6 in this case) and using Sn-(Sn-1) - 1/2*(Sn-2) - (Sn-3)/6 = 0, where Sn= a^n + b^n + c^n.

[@davidfranklin2731]

Agreed. In fact, it's probably slightly easier still to not actually find abc but instead leave the constant term in the cubic as an unknown constant D, then run the recurrence once to find a^3+b^3+c^3 in terms of D (and hence determine D).

[@ayanjain3106]

I too did with the same method .

[@goup867]

@Cool Dude I am from 10th grade too so they probably may :)

[@hunghinsun2123]

Actually, there is no need to consider the trinomial expansion. In order to achieve the result of abc=1/6, we can just consider the simple equation of (a+b+c)(a^2+b^2+c^2)=2, with the help of the previous result of ab+bc+ca=-1/2 and the given conditions a+b+c=1 and a^3+b^3+c^3=3.

[@nilsh5027]

Glad to see someone else used the same method! In case anyone else was curious, here's a little more detail: Given a + b + c = 1 and a^2 + b^2 + c^2 = 2, we can write: 2 = (a + b + c)(a^2 + b^2 + c^2) = a^3 + a(b^2) + a(c^2) + (a^2)b + b^3 + b(c^2) + (a^2)c + (b^2)c + c^3 = a^3 + b^3 + c^3 + ab(a + b) + bc(b + c) + ac(a + c) = a^3 + b^3 + c^3 + ab(1 - c) + bc(1 - a) + ac(1 - b) = a^3 + b^3 + c^3 + ab + bc + ac - 3abc Given ab + bc + ac = -(1/2) and a^3 + b^3 + c^3 = 3, we have: 2 = 3 - (1/2) - 3abc Which yields abc = 1/6

[@aplcc323]

Very cool, I made several mistakes trying to solve it by head. Thank you for showing me the errors in my ways!!

[@milankovacs4259]

I loved the video, explained really well and was extremely interesting to watch throughout.

[@ab45163j]

You're a champion!

[@togbot3984]

8:16 that's literally me. I do always solve algebra and get a unexpected result.

[@armanavagyan1876]

Pleasemore videos like this one thanks)

[@aidanstrong1061]

This question more or less came up in the 2009 STEP III paper for Cambridge maths admissions, except you had to also find a more general statement for it. Pretty cool!

[@cblpu5575]

Three years late, but there is a very simple and elegant way. We are given a+b+c = 1 Also a^2+b^2+c^2=2 =>(a+b+c)^2-2(ab+bc+ca)=2 but a+b+c=1 Therefore solving for ab+bc+ca, we have ab+bc+ca = -1/2. Now we can say that a,b,c are roots of the equation x^3-x^2-x/2+q=0 for some constant q. Using vietas relation we obtain that A_n-A_(n-1)-(1/2)(A_(n-2))+q(A_(n-3)) = 0 where A_n=a^n+b^n+c^n. Putting n=3 in the equation and solving we get q =-1/6. Now to get A_5 we need A_4. GeneratingA_4 using the recurrence relation gives us A_4=25/6. Generating A_5 the same way we get A_5=a^5+b^5+c^5=6

[@savaesmek]

cant you just multiply the relations like a^2+b^2+c^2 = 2 a^3+b^3+c^3 = 3 --------------------------------- (x) a^5+b^5+c^5 = 3x2 = 6

[@cblpu5575]

@@savaesmek no, you cannot. Even though in this case it yields the right solution, this is merely a coincidence. Are you familiar with the distributive property of multiplication? It states that a(b+c) = ab+ac. For example, 3(5+2) = 3(7) = 21 but also 3(5+2)= 3(5)+3(2)= 15+6=21. So, if you multiply the relations, you get (a^2+b^2+c^2)(a^3+b^3+c^3) = 6. From the distribute law,we have a^2(a^3+b^3+c^3)+b^2(a^3+b^3+c^3) + c^2(a^3+b^3+c^3) = 6 Using the distributive law on each of the terms again, you get a rather messy and long expression that is a ^ 5 + a ^ 3 * b ^ 2 + a ^ 3 * c ^ 2 + a ^ 2 * b ^ 3 + a ^ 2 * c ^ 3 + b ^ 5 + b ^ 3 * c ^ 2 + b ^ 2 * c ^ 3 + c ^ 5 = 6 As you can see, we do have the a^5, b^5 and c^5 terms. However there are a lot of other terms which we dont want. Ofcourse this can be solved this way with algebra and clever use of the right identities, however this is lengthy, and messy making you prone to maling mistakes.

[@savaesmek]

@@cblpu5575 yeah you're right im 14 and i just had an exercise at school where i had to add up relations so i thought you could do the same but i completely forgot what monstrosity of an answer i would get by multiplying

[@cblpu5575]

@@savaesmek haha, happens all the time

[@savaesmek]

@@cblpu5575 also since you were kind enough to respond could you please tell me what the branch of math with functions and graphs is called? i would like to learn about it before i get to that lesson

[@bot831]

imagine that the camera werent recording

[@_s_u_b_rah_manyam]

Real professor's patience shown here 👏👏👍🏻

[@givaru-tan9623]

Honestly I don't even have any interest in math but seeing this guy explain everything got me hooked on his lecture Weird

[@ikhu_nim]

a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) You can also find abc when you using this. And it's easy!

[@socerdemon8]

Did you ever build the pascal's pyramid? In wiki it says you build from above layers, which you didn't do.

[@mohamedabdullahi5708]

This is the only long math video that I have ever seen. He is hard worker proffesor.

[@othmaneasmar553]

Thank you very, i was watching the whole video without borrowing and that’s rare

[@benhetland576]

Well, well, well, at 38:00 we also know from numberphile that -1/12 = 1+2+3+4+5+..., so it doesn't cancel out against the 1/12 toward the end. Therefore the answer is negative infinity, isn't it? :-)

[@Logay495]

No

[@Logay495]

That's wrong

[@shailesh_rajpurohit]

16:00 he again write k in place of c😅😅😁

[@satyapalsingh4429]

Very Nice Method Of Teaching .Stay Blessed .

[@gregoriousmaths266]

NOW THAT WAS EPIC!!!!

[@Catishcat]

... I only realised I just watched 40 minutes of algebra after the video finished Math is immersive. Can't believe it was 40 minutes...

[@blackpenredpen]

: ))))))

[@bardozan]

Can't believe it was only 40 minutes. Would watch for a couple of hours.

[@jofx4051]

20 minutes if you 2X it

[@piept4651]

I raised (a+b+c)^5 and got interested about the relationship between dimensions (I just call it dimensions now, distinct variables within the parantheses, like this) and the power it is raised to the amount of each term in the expanded result, but I am too lazy to research it, so I just watch the video. Watched part of the video: I actually got 1. abc = 1/6 2. ab + bc + ca = -1/2 but didn't know what to do with it lol. And I got 2. from (a+b+c)(a^2+b^2+c^2) and (a+b+c)^3 instead.

[@blackpenredpen]

: ))) I didn't want to do that since (a+b+c)^5 is going to be crazy

[@nahian3454]

this guy’s really good at algebra

[@ethandavies953]

Much easier approach: First get out: a + b + c = 1 ab + bc + ac = -½ abc = ⅙ then define the sequence x_n=a^n+b^n+c^n x_n can be written as a recursive equations x_(n+3)-(1)x_(n+2)+(-1/2)x(n+1)-1/6 x_n =0 since the substitution gives x_n=z^n gives us the cubic with roots a,b,c. We know x_1, x_2, x_3 and then keep plugging in to get x_4 and x_5

[@heribertobarahona7695]

3:09 On the seventh term it should be a^3*c^2 Even though on the rest of the process you use the correct term, I just want to metion it because when you wrote a^2*c^2(a+c) on 5:08 I was getting confuse because I wasn't doing the excersie just looking, I was just looking at the terms you have.

[@ssdd9911]

using the same mathod, a^4+b^4+c^4=25/6

[@matteodamiano6733]

Thats almost 4

[@ammarbarbhaiwala9908]

Actually an easy way to do this is to make a third degree polynomial dx^3+ex^2+fx+g=0 having its roots as a, b, c and then using the properties of sum of roots and product of roots we can find a^4+b^4+c^4 this way this question can be done in less time.

[@fiNitEarth]

@@ammarbarbhaiwala9908 wie zum fick kommt einer Mit mois Profilbild unter so ein mathe Video? XD

[@AndDiracisHisProphet]

@@fiNitEarth Was für ein Typ? Das ist JC Denton

[@badereric]

@@AndDiracisHisProphet there is no god

[@aissofoxad]

the way you arranged the problem in 4:12 reminded me of symmetric matrix diagonalization question. Preety sure its just a coincedence or smth. Great video btw.

[@imc-indiangamer6265]

Please make more videos on algebra. A to Z.

[@backtothechickens]

Opens video with "lets do an extereme algebra question for fun". This is the youth culture I have been looking for.

[@jianxiang]

Even though I haven’t do the math, but I can imagine more symmetric formulas of x y z equal to an integer or simple fraction numbers, while they themselves are actually “monstrous” complex numbers. Just like an ice cream machine, with the complex mechanism inside hidden from you, it only spits out nice little ice creams😊

[@nocturnalvisionmusic]

Looks so simple but complex! 🤯😮😁

[@KCKaranChoudhary]

Whenever i want to sleep i watch your video, thanks bro!😀

[@aritrabandyopadhyay2706]

Instead of going for trinomial expansion, you could have gone for the relatively easy formula.... a³+b³+c³-3abc = (a+b+c)(a²+b²+c² - ab - bc - ac) and simply find abc from this

[@blackpenredpen]

Techie5879 nice formula!

[@raulvalgar1999]

The only man who could defeat Thanos

[@sostratuszgrimmagu]

Your passion for math is really catching😂

[@kimulo]

J'ai adoré merci ! Thank you !

[@ssdd9911]

-1/12!

[@gregoritsen]

Minus 1/12 factorial? Oh boy

[@saranshsoni171]

@@gregoritsen ikr

[@maxsch.6555]

approximately 1.055546564813466302313701484734262209810379114858428728414...

[@mosherosenthal4321]

Ikr

[@christiankameranbehnam6214]

@@maxsch.6555 that is more precise than pi

[@vilius...]

37:50 should have totally substituted the sum of all positive integers :^)

[@worldofclash5293]

U r explaination is good bro

[@ilyanebraij8495]

WoW that was extreme haha good video as always

[@mohdaftab2819]

I tried this question with newton's identities by making a polynomial with roots a b and c and got the right answer

[@nuzhatjahan4781]

That's what I was looking for

[@anmoljhamb9190]

Yeahhh! That's what I did too! It helped me solve the question in like 3 minutes or something. It even helped me get the values of a, b, and c!

[@aav56]

"Who tf would watch a 40 minute video of a guy doing algebra???" -Me, 40 minutes ago

[@78anurag]

Haha same

[@user-nd7df8gm6z]

For so many times when you are on a math competition but you are running out of time so you just took a most obvious guess for those kind of questions, and surprisingly there’s a great chance you are gonna get it right lol, simple answer of extreme complexity

[@crowdthrone6793]

Dude I’m not a very good problem solver in Algebra but this was too easy but blackpenredpen I liked the question!