A Tricky Roots of Cubics Problem

Given that α, β, γ are roots of the equation x^3 + 2x + 1 = 0, we calculate α^5 + β^5 + γ^5, using some elegant algebraic techniques and shortcuts.
00:00 Simultaneous equations
02:46 Vieta's formulas
04:55 Sum of squares
06:14 Sum of cubes
11:48 Sum of fifth powers

[@italyball2166]

There's also a nice side result here: the sum of the squares is negative, hence two of the roots are complex and conjugated

[@harry_dum7721]

Well the conjugate is a given since a polynomial cannot have a complex solution without it's conplex conjugate also being a solution.

[@abbeleon]

​@@harry_dum7721a polynomial can have non-real solutions that are not conjugate pairs if the coefficients are non-real

[@Errenium]

​@@harry_dum7721a polynomial over the reals

[@ivanerofeev1269]

Since α³=-2α-1, then α³+β³+γ³=-2(α+β+γ)-3=-3

[@DrBarker]

Nice, this saves us a lot of effort!

[@thatapollo7773]

Multiply by alpha^2 and you can do the same to obtain the 5th powers

[@wesleydeng71]

2:30, by the same token α^3+β^3+γ^3 = -2(α+β+γ)-3, which simplifies the calculation further.

[@danielmilyutin9914]

Yeah. I also noticed that. From other hand, we got some identity as byproduct.

[@erikr007]

To find the sum of the n-powers of the roots, just divide the polynomial x^n by x^3+2x+1. The remainder will be a quadratic so it will express α^n in terms of 1, α and α^2. Hence you'll only have to compute α+β+γ and α^2+β^2+γ^2 for any power of n.

[@gavintillman1884]

I’d do exactly what you did for alpha^2 + beta^2 + gamma^2 but there is a slicker approach for alpha^3 + beta^3 + gamma^3 using the original polynomial. alpha^3 = -2 alpha -1 and similarly for beta^3 and gamma^3 so the sum of those cubes is therefore -2(alpha + beta + gamma) -3 = -3.

[@alexlee6557]

P = a + b + c; Q = ab + bc + ac; R = abc; a^5 + b^5 + c^5 = P^5 - 5(P^2 - Q)(PQ - R)

[@HoSza1]

out of nowhere backwards derivation that shows none of the work you did, only the result...

[@mephist43]

@@HoSza1say that to Gauss and Ramanujan

[@HoSza1]

​@@mephist43Did you see any of them appear here? 😮

[@fahimuddin4401]

Everyone has given really slick solutions for sum of cubes, here is my way😅 begin by noticing alpha+beta+gamma=0 thus the sum of the cubes of the roots will be 3(alpha) (beta) (gamma) = 3(-1) = -3

[@jeanf6295]

You can define the recursive sequence relation : u(n+3)+2u(n+1)+u(n) = 0 Sequences that verify this relation have a closed form of the form : u(n) = A a^n+B b^n + C c^n where a,b and c are the roots of x^3+2x+1 (inject u(n) = x^n into the relation to see why) The specific sequence we are interested in is : u(n) = a^n + b^n + c^n (A=B=C=1) To compute u(n) using the recursive relation we need a set of three values : u(0) = 3 u(1) = a+b+c = 0 (as you have shown) u(2) = a²+b²+c² = (a+b+c)²-2(ab+bc+ca) From there use u(n+3) = -2u(n+1)-u(n) and you can get all values of u(n) easily.

[@danielmilyutin9914]

I love this approach!

[@MyOneFiftiethOfADollar]

If you are masochistic try the multinomial expansion on (alpha + beta + gamma)^5 The given cubic implies alpha + beta + gamma=0, (alpha*beta + alpha*gamma + beta*gamma)=2, alpha*beta*gamma=-1 (These are called elementary symmetric polynomials in the roots that are expressible as coefficients(Vieta) Then use (alpha + beta + gamma)^5 = 0 and some tedious algebra/factoring on the expanded form right hand side. I did not work it all out, but it is probably about the same amount of effort as you presented in video.

[@shacharh5470]

I did it a different way. I started with something you did too: a+b+c=0 ab+ac+bc=2 abc=-1 From the first equation c=-(a+b) (let's call this A) substitute that into the other equations you get: a^2b + ab^2 = 1 (B) and ab-(a+b)^2 = 2 (C) so using (A) and the binomial theorem, c^5 =-(a+b)^5 = -(a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5) so a^5 + b^5 + c^5 = -(5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4) = -5ab(a^3 + 2a^2b + 2a^b^2 + b^3) simplify this using (B) to = -5ab(a^3 + b^3 + 2) now use the binomial theorem again, but for the cube of sum: = -5ab((a+b)^3 -3(a^2b+ab^2) + 2) use (B) again: = -5ab((a+b)^3 -3 + 2) = -5ab((a+b)^3 -1) = 5(ab - ab(a+b)^3) use (B) once more: = 5(ab - (a+b)^2) now use (C): = 5(2) = 10

[@marcgriselhubert3915]

Let's note a, b, c the 3 roots of the equation (I have no greek letters) As a^3 +2.a +1 = 0, we have a^3 = -2.a -1 and a^5 = -2.a^3 -a^2 = -2.(-2.a -1) -a^2 = -a^2 +4.a +2 Idem with b and C So, a^5 + b^5 + c^5 = -(a^2 + b^2 + c^2) +4.(a + b + c) +6 Now a + b + c = 0 and a.b + a.c + b.c = 2 (symetric functions of the roots of the equation) So (a^2 + b^2 + c^2) =(a + b + c)^2 - 2.(a.b +a.c + b.c) = 0^2 - 2.2 - -4 Finally: (a^5 + b^5 + c^5) = - (-4) +4.0 + 6 = 10

[@user-we6zx5tw7d]

Perhaps the shortest solution is to calculate a2 + b2 + c2 by using Vieta’s formula (I replaced the roots with a b c and x2 means a square) and a3 + b3 + c3 by replacing the roots in the equation. Another interesting approach (because it starts from nowhere) is the following: Let’s define X= a4 + b4 + c4 and let’s multiply this equation by a+b+c (=0 by Vieta’s formula) 0= a5 + b5 + c5 + a*(b4+c4) + b*(a4+c4) + c*(a4+b4); now by multiplying out and rearranging the summands we get (*) a5 + b5 + c5 = -ab*(a3+b3) - bc*(b3+c3) - ac*(a3+c3); Let’s transform any of the symmetric summands on the right by using the formula for the sum of cubs: ab*(a3+b3) = ab*(a+b)*(a2-ab+b2) = ab*(a+b)*((a+b)2 - 3ab)= {by using abc = -1 and a+b+c=0 we have ab= (-1)/c and (a+b)= (-c)} = c2 - 3ab Replacing these result in (*) we receive a5 + b5 + c5= -c2 + 3ab - a2 + 3bc - b2 + 3ac, by adding and subtracting 2ab + 2bc + 2ca and applying the formula for the sum of the squares we finally get a5 + b5 + c5= -(a + b + c)2 + 5(ab + bc + ca)= 0 + 5*2 = 10

[@jursamaj]

Just a note for the future: standard text notation for powers is x^p, so x cubed is x^3. Or, if you have something that inserts unicode for you, superscripts are easy: x³. On Windows, the Character Map app is builtin, while on Mac, the 'keyboard and emoji viewer' menu does it.

[@supasayajinsongoku4464]

This is the first youtube problem ive ever solved

[@holyshit922]

Symmetric polynomials In fact we have special case called power sums Newton-Girard formulas then Vieta formulas should solve the problem To use Vieta formulas you must have elementary symmetric polynomials and Newton-Girard formulas allow to express power sums in terms of elementary symmetric polynomials

[@ahoj7720]

Symmetric polynomials and logarithmic derivatives (hence Newton-Girard formulas) are very powerful tools.

[@television-channel]

that was kinda thrilling

[@user-cd9dd1mx4n]

Thank you very much for such great video. I know many great books in algebra, containing such problems. But I am looking for a better one in your opinion, could you please suggest? Thanks❤

[@easymathematik]

A very nice approach is to use polynomial long division involving the function and its derivative

[@nasrullahhusnan2289]

Let a, b, c be roots of x³+2x+1=0 Then: • a+b+c=0 • ab+bc+ca=2 • abc=-1 Note that • 0=(a+b+c)² =a²+b²+c²+2(ab+bc+ca) =a²+b²+c²+2×2 --> a²+b²+c²=-4 • 0=(a+b+c)³ =a³+b³+c³+3(ab+bc+ca)(a+b+c) -3abc =a³+b³+c³+3 --> a³+b³+c³=-3 • 12=(a³+b³+c³)(a²+b²+c²) =a⁵+b⁵+c⁵+a³b²+a³c² +b³a²+b³c²+a²c³+b²c³ =a⁵+b⁵+c⁵+a²b²(a+b+c)-a²b²c +b²c²(b+c+a)-ab²c² +a²c²(c+a+b)-a²bc² =a⁵+b⁵+c⁵-a²b²c-ab²c²-a²bc² =a⁵+b⁵+c⁵-abc(ab+bc+ac) =a⁵+b⁵+c⁵+2 Therefore a⁵+b⁵+c⁵=10

[@thatapollo7773]

A simpler approach is note that f(alpha) + f(beta) + f(gamma) = 0, which gives us the sum of cubes and then note alpha^2 f(alpha) + beta^2 f(beta) + gamma^2 f(gamma)=0 which gives us the 5th powers sum

[@Khashayarissi-ob4yj]

With luck and more power to you.

[@wannabeactuary01]

Let roots be a, b, c Noting abc = -1 , ab +bc+ ca = 2 , a + b+ c = 0, a^2 + b^2 + c^2 = (a+b+c)^2 - 2(ab + bc + ca) = -4 And a^3 = -2a - 1 ⇒ a^5 = (a^2)(-2a -1) = -2a^3 -a^2 I) a^5 = -a^2 +4a + 2 II) b^5 = -b^2 +4b + 2 III) c^5 = -c^2 +4c + 2 -------------------------------------- Adding (a^5 + b^5 + c^5) = -(-4) + 4(0) + 6 = 10

[@spacer999]

Rewrite the eqn as x^3 = -2x-1 to easily solve α^3+β^3+γ^3=-2(0)-3=-3 since α+β+γ=0 Rewrite the eqn as x^2 = -2-(1/x) to easily solve α^2+β^2+γ^2=-6-(2/-1)=-4 since αβ+αγ+βγ=2 and αβγ=-1

[@honestadministrator]

Another way x^3 + 2 x + 1 = 0 implies alpha + beta + gamma = 0 alpha * beta + beta * gamma + gamma * alpha = 2 x^5 + 2 x^3 + x^2 = 0 x^5 = 2 ( 2 x + 1) - x^2 Hereby alpha ^5 + beta ^ 5 + gamma ^5 =- alpha ^2 -beta ^ 2 - gamma ^5 + 4 ( alpha + beta + gamma) + 6 = - ( alpha + beta + gamma) ^2 + 2 (alpha * beta + beta * gamma + gamma * alpha) +6 = 2 * 2 + 6 = 10

[@shogun6943]

If a+b+c=0 then a³+b³+c³=3abc or in this case alpha beta and gamma(i couldnt find them in my keyboard)

[@shogun6943]

It works for all numbers btw

[@ferronzomeren2733]

4:52 I want to try to solve this system of equations.. I will post the results as a reaction

[@ferronzomeren2733]

It does not help, as you end up with another cubic equation. In that case you can just calculate the original roots and you are done as well

[@johnstanley5692]

Easier? let d(x)=x^3+2*x+1=0 => x^5 = -x^2 +4*x+2 (i.e. remainder from x^5/d(x)). Given a+b+c =0 => a^5+b^5+c^5 = 6-(a^2+b^2+c^2). Now (a+b+c)^2=a^2+b^2+c^2+2*(a*b+a*c+b*c). now (a*b+a*c+b*c)=2 > a^5+b^5+c^5 = 6-(-4) = 10.

[@thiagoflaherty9992]

Can You prove that √((100!×99!×98!.....4!×3!×2!)50!) is a integer?

[@datokvartskhava4711]

x^3+2x+1=0 does not have 3 solution it ahs only one if you graph the function

[@jamesstrickland833]

it has 1 real solution but two complex solutions.

[@samyachakraborty263]

who needs netflix

[@TikeMyson69]

Math vids and chill

[@worldnotworld]

Way too fast! I didn't even know it was over!

[@golddddus]

1) α³ + β³ + γ³ + 2(α + β + γ) + 3 = 0 tanks @ivanerofeev. 2) x^2 + 2 + 1/x = 0 α ^2 + β^2 + γ^2 + 6 + (αβ + αγ +βγ)/αβγ = 0 α ^2 + β^2 + γ^2 + 6 + (2)/(-1) =0 α ^2 + β^2 + γ^2 = -4 😎