Cuberoot(2+ sqrt5) + Cuberoot(2 - sqrt5)

[@kjaideep]

Sir a teacher like you in this generation inspires us to go deeper into the concepts and tries to know the beauty of it

[@BartBuzz]

Another clever yet simple solution. I can see why so many who are new to math enjoy your channel. Your approaches and explanations even bring back fond memories of my post-grad days so many years ago.

[@rabotayet]

I am from Brazil. I do not speak english. Your explanations are so clear that we do not need to speak any language to understand your explanations. Congratulations. I am 75.

[@PrimeNewtons]

Thank you

[@CaioSilva-ph3ro]

Concordo. A didática é tão boa que a diferença de línguas nem importa

[@Faroshkas]

Nunca parei para pensar por esse ângulo. O melhor professor consegue ensinar para aquele que nada sabe.

[@luisclementeortegasegovia8603]

You're the architect of substitutions 👍 hard to see at one look! Loved this solution, thanks!

[@Grassmpl]

Pretty good comment for someone that does not speak English.

[@skwbusaidi]

Second method for solving c^3+3c-4 =0 is to split the constant: c^3-1 +3c-3=0 (c-1)(c^2+c+1)+3(c-1)=0 (c-1)(c^2+c+4)=0 Third method When you get a polynomial equation first to check the sum of the coefficient and if it is 0 the x=1 which is true hare and we do long or synthetic division ( info : another one to check the sum of the even power coefficient and the sum of the odd power coefficien; and if they are equal., then x=-1 is solution)

[@trkncrgl]

Never stop learning!

[@crowbar_the_rogue]

6:30 This guy can read minds.

[@childrenofkoris]

SO SATISFYING WATCHING THIS... THANK YOUUUU

[@arsessahra-yb9zb]

افرین راه حل بسیار جالبی بود سپاس از شما

[@YAWTon]

Yes, interesting indeed.

[@prakashchandradosi6115]

Hats off🎉 excellent person who can make the world math_loving society 🎉

[@dfontoul]

This is brilliant. Well done sir.

[@dan-florinchereches4892]

Nice problem... My tought initially was that we have square root if 4 and 5 under cube root... just 1 off eachother. Then i wondered if the top is a cubic epansion then we need to have at least a 3√5+5√5=8 so multiplying and dividing by 8 gave a nice perfct cube. Curious about your approach. Your song from channel intro is so good and your voice and approach are so calming. Can you recommend some books regarding pedagogy or didactics?

[@AzmiTabish]

Be curious like a child and learning becomes fun 🙂

[@ProactiveYellow]

By the definition of the cube root function (which is a FUNCTION) there is only one solution. The common convention is that ³√x only takes the real solution to z³=x, even though there are three solutions in the complex plane. Remember that the definition of a function requires that every member of the domain maps to only one member of the range (in other words you can only get one output for every input). We know that our answer is real and positive by approximation (2+√5 is greater than 1, 2-√5 is between 0 and 1, so their cube roots will have the same relationship), so when we solve for the cubic in C, we find only one real solution, which must be our answer, as the other solutions are introduced extraneous solutions from cubing the whole thing.

[@Amoeby]

You meant cbrt(2 - sqrt(5)) is in the range from -1 to 0, not from 0 to 1, right?

[@ProactiveYellow]

@@Amoeby yes. I meant the absolute value, which then keeps the whole thing positive when subtracting the negative part from the positive part.

[@emadmohammed2054]

thank you very much Mr for all things that I've learned from you...

[@2106522]

It was amazing! 👍 Thank you very much!

[@mcumer]

I solved it in a similar way: ( a+b)^3= a^3+b^3+3ab(a+b).. therefore, as the teacher does, I have elevated the sum of root to the third power, obtaining 4-3x, where x is the sum of the cubic roots.. the equation is x^3= 4-3x, that is x^3+3x-4=0.. this equation has on obvious solution which is 1.. so we obtain that the sum is 1.. Another solution: the sum of cubic roots reminds me the formula for the equation x^3+px+q=0.. Comparing the 2 formulas we obtain -q/2=2 and q^2/4+p^3/27=5.. from the first one we obtain q=-4. If we put this value in the second one, we obtain p^3/27=1 that implies p=3 .. the rest of the demonstration is egual

[@Algebrian_Guy]

This is the only person after Bob Ross who goes by the saying "Trust the process"

[@Tshego2000]

these problems make me feel stupid, but also encourage me to better my math skills.

[@samirayoub6100]

I was not good all times in mathematics as a student, but know I want to understand how mathematics help in life. I enjoy your learning. But in this exercise, I don't understand the importance of complex solutions. As you said, we can stop at c =1, that's what gives a calculator ! The cubic root of a negative irrational number is an irrational number.

[@Amoeby]

As soon as I saw this it reminded me of the answer to the cubic equation using Cardano formula. So what you've done is basically reversed that process by recreating that cubic and then factorised it. Though, those complex solutions aren't our answer because you got them after you cubed both sides.

[@Amoeby]

@@NadiehFan don't want to disregard your effort but I'm not sure if it was necessary to write the whole thing down with explanations considering that I wrote about Cardano formula which implies that I know how it works.

[@Faroshkas]

Hi! I don't understand how the expression can have imaginary solutions. Isn't the cube root of a real number always real? And therefore, the sum of two of them also real? Amazing video! You are an amazing teacher. ❤

[@NadiehFan]

The expression ∛(2 + √5) + ∛(2 − √5) does not have multiple values. Its value is _one_ of the solutions of the cubic equation c³ + 3c − 4 = 0 and since the expression to evaluate consists of the sum of two cube roots of real numbers which are each real their sum must also be real. Unless stated otherwise, the cube root of a real number is always assumed to be its unique real value. But what about the two other (complex) solutions of the cubic equation? Well, Prime Newtons started with ∛(2 + √5) = a ∛(2 − √5) = b and c = a + b Now, as shown in the video, c = a + b is a solution of the cubic equation c³ + 3c − 4 = 0 so we have (a + b)³ + 3(a + b) − 4 = 0 But we also have the identity (a + b)³ = (a³ + b³) + 3ab(a + b) which can be written as (a + b)³ − 3ab(a + b) − (a³ + b³) = 0 and comparing this with our equation (a + b)³ + 3(a + b) − 4 = 0 we can see that a and b must satisfy ab = −1 a³ + b³ = 4 This is of course no surprise, because you can directly verify this by starting from a = ∛(2 + √5), b = ∛(2 − √5). As I explain in another comment on this video, both ∛(2 + √5) and ∛(2 − √5) can be denested and we can prove that we have ∛(2 + √5) = ½ + ½√5 ∛(2 − √5) = ½ − ½√5 So, we have a = ½ + ½√5 b = ½ − ½√5 as a real solution pair (a, b) of our system ab = −1 a³ + b³ = 4 But what about other solutions of this system? We know that for any pair (a, b) that is a solution of this system c = a + b is a solution of the cubic equation c³ + 3c − 4 = 0, so there must be two other solution pairs (a, b) of our system in a and b. These other solutions pairs are not hard to find. Letting (a₁, b₁) = (a, b) be the real solution pair we only need to multiply the real values a = ½ + ½√5 and b = ½ − ½√5 with the complex cube roots of unity ω₁ = −½ + i·½√3, ω₂ = −½ − i·½√3 in either order to find our other two solution pairs (a₂, b₂) = (ω₁a, ω₂b) and (a₃, b₃) = (ω₂a, ω₁b). Since there are three different values for a and three different values for b you might be tempted to think that we can form 9 solution pairs of our system of equations in a and b. But of course this is not true since (a₁, b₁), (a₂, b₂), (a₃, b₃) are the only three pairs that satisfy ab = −1 because ω₁ω₂ = 1 whereas ω₁² = ω₂, ω₂² = ω₁. Since each solution pair (a, b) gives a solution c = a + b of the cubic equation c³ + 3c − 4 = 0 we find the three roots of this equation which are c₁ = a₁ + b₁ = a + b c₂ = a₂ + b₂ = ω₁a + ω₂b c₃ = a₃ + b₃ = ω₂a + ω₁b and this gives c₁ = (½ + ½√5) + (½ − ½√5) = 1 c₂ = (−½ + i·½√3)(½ + ½√5) + (−½ − i·½√3)(½ − ½√5) = −½ + i·½√15 c₃ = (−½ − i·½√3)(½ + ½√5) + (−½ + i·½√3)(½ − ½√5) = −½ − i·½√15

[@nathanaelhahn]

All numbers have n nth roots, most of which are usually complex-- so the original expression does indeed have nine meanings, assuming we allow for complex cube roots. Usually, though, we mean the real root when we use the radical symbol.

[@ManjulaMathew-wb3zn]

Great explanation!

[@seekingCK]

nice how you just made the cube root of -1 into just -1 not using complex numbers. fun problem

[@jussihamalainen7692]

By trial, I found that (1+sqrt(5))/2 cubed equals 2 + sqrt(5). From that, it was easy to guess that (1-sqrt(5))/2 cubed is 2 - sqrt(5). Add them and you get 1.

[@vafasadrif12]

Well actually you get 4 when you add them🤓🤓

[@simoneeeq6421]

great video to be honest

[@AbhinavJaat-fw1fc]

Sir I am from India but still your big fan

[@robidfen5459]

I have seen something similar and I think it is fair to say that ∛(a + √(a²+1)) + ∛(a - √(a²+1)) = 1 for every a

[@DefenderTerrarian]

Unfortunately, that is not the case. The equation you have proposed only =1 when a=2.

[@robidfen5459]

​@@DefenderTerrarianOk, I didn't check it, you are probably right

[@NadiehFan]

@@robidfen5459 Well you should have checked it of course before posting. If we have the _equation_ (not identity) ∛(a + √(a²+1)) + ∛(a − √(a²+1)) = 1 and we want to solve this for a, we can cube both sides and apply the identity (p + q)³ = p³ + q³ + 3pq(p + q) with p = ∛(a + √(a²+1)), q = ∛(a − √(a²+1)) to the left hand side to obtain (a + √(a²+1)) + (a − √(a²+1)) + 3·∛((a + √(a²+1))(a − √(a²+1)))·(∛(a + √(a²+1)) + ∛(a − √(a²+1)) = 1 Looks daunting, but we can simplify this a great deal because (a + √(a²+1)) + (a − √(a²+1)) = 2a, (a + √(a²+1))(a − √(a²+1)) = a² − (a² + 1) = −1 and of course ∛(a + √(a²+1)) + ∛(a − √(a²+1)) = 1 as per the original equation, so we end up with 2a − 3 = 1 which has only one solution which is a = 2

[@RyanLewis-Johnson-wq6xs]

Cbrt[2+Sqrt[5]]+Cbrt[2-Sqrt[5]]=1 (-1 ± Sqrt[15]i)/2 final answer

[@vasanthalakshmi6734]

One doubt sir 'C' have only numbers which constant so addition of constants must be a constant ?

[@EvilSandwich]

Be honest, you made that problem specifically so that the complex solutions would be the golden ratio. I only asked that because in hindsight, that's totally what I would have done LOL

[@PrimeNewtons]

This was a Team Selected Test for IMO in 2009. I didn't make it up.

[@EvilSandwich]

@@PrimeNewtons Haha good to know. I wonder if they did that on purpose.

[@pauselab5569]

denesting is pretty difficult though you could use the brute force approach each time (simply nth power everything). sometimes if there are special forms, it can be simplified. I know this by heart because it leads to the golden ratio and its conjugate(very famous). so the sum is just 1.

[@NadiehFan]

There actually is a standard procedure for denesting individual denestable cube roots of quadratic surds like ∛(2 + √5). Assume there exist _rational_ numbers x and y with √y _irrational_ such that (1) ∛(2 + √5) = x + √y then we must also have (2) ∛(2 − √5) = x − √y From (1) and (2) we have x² − y = (x + √y)(x − √y) = ∛(2 + √5)·∛(2 − √5) = ∛((2 + √5)(2 − √5)) = ∛(2² − (√5)²) = ∛(4 − 5) = ∛(−1) = −1, so (3) x² − y = −1 Also from (1) the cube of x + √y must equal 2 + √5 so we have (x + √y)³ = x³ + 3x²√y + 3xy + y√y = (x³ + 3xy) + (3x² + y)√y = 2 + √5 which implies (4) x³ + 3xy = 2 From (3) we have y = x² + 1 and substituting this in (4) we get x³ + 3x(x² + 1) = 2 which gives (5) 4x³ + 3x − 2 = 0 Since x must be rational, we are looking for rational solutions of this cubic equation. According to the rational root theorem, for any potential rational solution m/n or −m/n of (5), m must divide the absolute value of constant term, which is 2, and n must divide (the absolute value of) the coefficient of the cubic term, which is 4, so m can be 1 or 2 and n can be 1 or 2 or 4. However, we can also see that (a) the polynomial on the left hand side of (5) is strictly increasing on the real number line and (b) the left hand side of (5) is negative for any negative x. This means, first of all, that (5) can only have a single real solution and, secondly, that any real solution and eo ipso any rational solution must be positive. So, we only need to test ¼, ½, 1 and then we quickly find that x = ½ is a solution and consequently the only real solution of (5). With x = ½ we have y = x² + 1 = (½)² + 1 = ¼ + 1 = ⁵⁄₄ and so we have ∛(2 + √5) = ½ + √(⁵⁄₄) which can also be written as ∛(2 + √5) = ½ + ½√5 Of course (1) and (2) imply that we must also have ∛(2 − √5) = ½ − ½√5 Reference: Kaidy Tan, Finding the Cube Root of Binomial Quadratic Surds. _Mathematics Magazine_ Vol. 39, No. 4 (Sep., 1966), pp. 212-214 (JSTOR 2688084)

[@JASONKINGMATHK]

Cool thanks

[@GreenMeansGOF]

This example demonstrates precisely why the cubic formula is not taught. The cubic formula will give us the two cube roots when the answer is simply an integer.

[@davidbrisbane7206]

Cardano's depressed cube root formula tells us that if x³ + px + q = 0, then c is a real root of this equation where c = u + v and p and q are real, such that u = ∛[-q/2 + √[(q/2)² + (p/3)³]] and v = ∛[-q/2 - √[(q/2)² + (p/3)³]] So, let 2 = -q/2, so q = -4. Let (q/2)² + (p/3)³ = 5, So, 4 + (p/3)³ = 5. So, p = 3. So, the cubic with root c = u + v is c³ + 3c - 4 = 0, which as the video shows, has a real root c = 1. This approach isn't guaranteed to work in the general case for adding two cube roots together, but it does work for the addition of the two cube roots of the form shown in this example.

[@zanti4132]

What's weird, though, is that when you apply Cardano's formula to get the real root of x³ + 3x - 4 = 0, you get as the real root the nested radical in the video, not 1. So how do you show this messy nested radical can be reduced to something simple, like an integer? You unravel the nested radical as done in the video, returning to the original cubic - in other words, you go back to Square One. Then you revert to the primitive approach you learned in high school to factor the cubic (this approach is shown in the video) to see if the cubic can, in fact, be factored, and there is no guarantee of the factorization working. Now, you can do a test that will tell you ahead of time if a nested radical in this format can be simplified. For the nested radical in the video, pull out the integers a = 2 and b = 5. If a² - b is a perfect cube, the simplification will work. If a² - b is not a perfect cube, don't waste your time. For the nested radical in the video, you get 2² - 5 = -1, which is a perfect cube, so it's all good.

[@NadiehFan]

@@zanti4132 Let a be a nonzero _rational_ number, b a positive _rational_ number with √b _irrational_ and suppose there exists a _rational_ number x as well as a positive _rational_ number y with √y irrational such that (1) ∛(a + √b) = x + √y then we must also have (2) ∛(a − √b) = x − √y Multiplying (1) and (2) we have (3) ∛(a² − b) = x² − y Now, the right hand side of (3) is rational, so the left hand side of (3) must also be rational which implies that a² − b is indeed the cube of a rational number. This means that a² − b being the cube of a rational number is indeed a _necessary condition_ for the denestability of ∛(a + √b) and ∛(a − √b). But the converse is _not true_ which means that a² − b being the cube of a rational number is _not sufficient_ to guarantee denestability of ∛(a + √b) and ∛(a − √b). Example: for ∛(3 + √8) we have a = 3, b = 8, so a² − b = 9 − 8 = 1 = 1³ is the cube of a rational number, but ∛(3 + √8) is _not_ denestable.

[@jamesmarshall7756]

Easy to find that : (1+ sqrt5)^3 = 2^3 (2+sqrt5) and (1- sqrt5)^3=2^3 (2-sqrt5). So after that, you get the answer because cuberoots disappear.

[@ПетрКурнев]

it's surprising that 2 + √5=(φ₁)³ , 2 - √5=(φ₂)³ , where φ₁=0.5*(1 + √5), φ₂=0.5*(1 - √5) are two solutions of the quadratic equation φ² = φ + 1 so, φ₁ + φ₂= 1 .

[@poonpoonz6024]

Can you do a modulo questions

[@albertbayetkingni8101]

Vous êtes un homme

[@bdofficiallitonlife]

Good

[@rikutitsoya5597]

Thank you teacher, but how can we prove it, it is a natural number ( without comput it )

[@NadiehFan]

He actually _did_ prove it: the value c of the expression to evaluate is real _and_ it is a root of the equation c³ + 3c − 4 = 0 _and_ the only real root of this equation is c = 1. It is also possible to prove that ∛(2 + √5) = ½ + ½√5 ∛(2 − √5) = ½ − ½√5 which of course gives ∛(2 + √5) + ∛(2 − √5) = (½ + ½√5) + (½ − ½√5) = ½ + ½ = 1 since the irrational parts of both cube roots cancel each other when they are added.

[@Abid_Ibn_Ashraf]

How can adding two real number gives us 1 real solution and two imaginary solutions? You cannot just say 2= a+bi where b≠0

[@justsomeboyprobablydressed9579]

2 + sqrt(5), for example, has three cube roots, two of them non-real. Someone might interpret the original expression as a set of complex values (nine, actually). However, when the cube-root symbol is wrapped around a real number, I think it is meant to be interpreted as a real number only. Also, the reason he got three complex solutions, instead of nine, happened when he only used the real cube root of -1. If you like, you can interpret each cube root as a real number as you suggest, and then you get only c = 1. Ignore the rest.

[@NadiehFan]

@cubification4823: You can't. The expression ∛(2 + √5) + ∛(2 − √5) is _one_ of the solutions of the cubic equation c³ + 3c − 4 = 0 and since the expression to evaluate consists of the sum of two cube roots of real numbers which are each assumed to be real their sum must also be real. Unless stated otherwise, the cube root of a real number is always assumed to be its unique real value. But what about the two other (complex) solutions of the cubic equation? Well, Prime Newtons started with ∛(2 + √5) = a ∛(2 − √5) = b and c = a + b Now, as shown in the video, c = a + b is a solution of the cubic equation c³ + 3c − 4 = 0 so we have (a + b)³ + 3(a + b) − 4 = 0 But we also have the _identity_ (a + b)³ = (a³ + b³) + 3ab(a + b) which can be written as (a + b)³ − 3ab(a + b) − (a³ + b³) = 0 and comparing this with our _equation_ (a + b)³ + 3(a + b) − 4 = 0 we can see that a and b must satisfy ab = −1 a³ + b³ = 4 This is of course no surprise, because you can directly verify this by starting from a = ∛(2 + √5), b = ∛(2 − √5). As I explain in another comment on this video, both ∛(2 + √5) and ∛(2 − √5) can be denested and we can prove that we have ∛(2 + √5) = ½ + ½√5 ∛(2 − √5) = ½ − ½√5 So, we have a = ½ + ½√5 b = ½ − ½√5 as a real solution pair (a, b) of our system ab = −1 a³ + b³ = 4 But what about other solutions of this system? We know that for any pair (a, b) that is a solution of this system c = a + b is a solution of the cubic equation c³ + 3c − 4 = 0, so there must be two other solution pairs (a, b) of our system in a and b. These other solutions pairs are not hard to find. Letting (a₁, b₁) = (a, b) be the real solution pair we only need to multiply the real values a = ½ + ½√5 and b = ½ − ½√5 with the complex cube roots of unity ω₁ = −½ + i·½√3, ω₂ = −½ − i·½√3 in either order to find our other two solution pairs (a₂, b₂) = (ω₁a, ω₂b) and (a₃, b₃) = (ω₂a, ω₁b). Since there are three different values for a and three different values for b you might be tempted to think that we can form 9 solution pairs of our system of equations in a and b. But of course this is not true since (a₁, b₁), (a₂, b₂), (a₃, b₃) are the only three pairs that satisfy ab = −1 because ω₁ω₂ = 1 whereas ω₁² = ω₂, ω₂² = ω₁. Since each solution pair (a, b) gives a solution c = a + b of the cubic equation c³ + 3c − 4 = 0 we find the three roots of this equation which are c₁ = a₁ + b₁ = a + b c₂ = a₂ + b₂ = ω₁a + ω₂b c₃ = a₃ + b₃ = ω₂a + ω₁b and this gives c₁ = (½ + ½√5) + (½ − ½√5) = 1 c₂ = (−½ + i·½√3)(½ + ½√5) + (−½ − i·½√3)(½ − ½√5) = −½ + i·½√15 c₃ = (−½ − i·½√3)(½ + ½√5) + (−½ + i·½√3)(½ − ½√5) = −½ − i·½√15

[@ThePullumFamily]

I would change the title to ³√(2 + √5) + ³√(2 - √5)

[@sinichitaniyama]

well dressed

[@brian554xx]

title says sqrt2 twice while the thumbnail shows sqrt5 each time. edit: fixed quickly!

[@ThenSaidHeUntoThem]

I see he fixed it

[@PrimeNewtons]

Thanks. I corrected it

[@rafaelsimoes2974]

I put in Wolfram Alpha and just show a complex number, why?

[@Rai_Te]

Wolfram Alpha wears no hat (scnr).

[@NadiehFan]

@rafaelsimoes2974 Impossible to say if you don't tell _exactly_ what you entered in WolframAlpha. If you enter cbrt(2 + sqrt(5)) + cbrt(2 - sqrt(5)) then you will get the correct result 1 immediately. So, yes, WolframAlpha can give the correct answer. Of course this assumes that we use _real_ cube roots. You can also opt to use the _principal_ cube root in WolframAlpha. The principal cube root of a _positive_ real number is equal to the real root of that number, but the principal cube root of a _negative_ real number is a complex number. In fact, the principal cube root of a negative real number is equal to the real cube root of its opposite multiplied by the principal cube root of −1 which is ½ + i·½√3. So, when you choose to work with the principal values of the cube roots, the result will be complex.

[@rafaelsimoes2974]

@@NadiehFan thank you, Wolfram has consider a complex value when i put (2+√5)^(1/3), cbrt returns real value

[@YAWTon]

That happens in other systems as well. E.g. in Python/sympy, the cube-root function cbrt(x) return the principal root y of the equation y^3=x. if x is negative the principal root is by definition a complex number. For the real root in sympy, one would have to use the real_root function, y=real_root(x,3).

[@johnstorey7547]

@@NadiehFan😂

[@azzahranazyaputricatappa6468]

but how the cubic root dissapear can anyone explain?

[@prime423]

Its obvious 1 is a root. Synthetic division makes it trivial.

[@antiXIII]

Keep it real

[@GPSPYHGPSPYH-ds7gu]

The Universe actual understanding is Mathematic PAZA M C69AoneA

[@holyshit922]

(10a+b)^2 = 100a^2+20ab+b^2 (10a+b)^2 - 100a^2 = 20ab+b^2 (10a+b)^2 - 100a^2 = (20a+b)b 1. We divide number into two digit groups from decimal point in both directions 2. Choose digit such that difference between the leftmost group and square of chosen digit is as close to zero as possible but nonnegative 3. Let call that difference remainder Append next two digits group to the remainder On the side double actual approximation , append last digit of next approximation and multiply number created in this way by last digit of next approximation Update remainder by calculating the differene between remainder and number calculated on side Next digit you choose in such way that difference between remainder and number calculated on the side is as close to zero as possible but nonnegative 4. Repeat step 3. until reminder is zero and there are no groups to append or you get enough digits of approximation (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 1. We divide number into three digit groups from decimal point in both directions 2. Choose digit such that difference between the leftmost group and cube of chosen digit is as close to zero as possible but nonnegative 3. Let call that difference remainder Append next three digits group to the remainder On the side calculate square of actual approximation and triple it , append square of last digit of next approximation to tis number then add triple product of actual approximation and last digit of next approximation shifted one position to the left then multply number created in this way by last digit of next approximation Update remainder by calculating the differene between remainder and number calculated on side Next digit you choose in such way that difference between remainder and number calculated on the side is as close to zero as possible but nonnegative 4. Repeat step 3. until reminder is zero and there are no groups to append or you get enough digits of approximation

[@creativename.]

Yay blackboard 🗣🔥

[@danobro]

Technically, there are nine possible solutions to this expression, as there are three complex answers to the first term, and three to the second.

[@jacobgoldman5780]

well you would need to check that the sum of all the pairs are distinct might have the same solution appear multiple times...

[@ingiford175]

It collapses to three I think

[@ingiford175]

Things in the form (a + b)^1/3 + (a-b)^1/3 where a is in Z and b is squareroot of a number in Z will collapse to a cubic instead of a 9th power.

[@danobro]

@@jacobgoldman5780 He just hasn't in this video. that's all

[@staskrivosheev1349]

Решение не правильное. Правильное решение =1. Автор судак.

[@comdo777]

answer=2/5

[@trebejoescaques6095]

May be the answer is wrong: c1 = [2 + Sqrt[5] ]^(1/3) = 1.618033989 c2 = [2 - Sqrt[5] ]^(1/3) = 0.3090169944 + 0.5352331347 i So c = c1+ c2 = 1.927050983 + 0.5352331347 i. Then, c is not a real number, is a complex number.

[@哲子仮免]

You are like Tambuwal.