A Tricky Roots of Cubics Problem

[@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.

[@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!

[@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.

[@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.

[@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

[@mathcanbeeasy]

a+b+c=0
ab+ac+bc=2
a^2+b^2+c^2=0^2-2*2=-4
The sum of cubes is obvious from the equation.
a^3+b^3+c^3=-2(a+b+c)-3=-3
Then, multiplying the equation with x^2 we have
a^5+b^5+c^5=-2*(a^3+b^3+c^3)-(a^2+b^2+c^2)=-2*(-3)-(-4)=6+4=10
In fact, for any S_n=a^n+b^n+c^n we don't need the expansion of (a+b+c)^n. Every S_n can be obtained by recuration from the equation and Viette.

[@television-channel]

that was kinda thrilling

[@supasayajinsongoku4464]

This is the first youtube problem ive ever solved

[@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❤

[@misterguts]

ChatGPT 4o solved this problem from the statement in the video description (slightly edited):
"Given that α, β, γ are roots of the equation x^3 + 2x + 1 = 0, calculate α^5 + β^5 + γ^5."

[@easymathematik]

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

[@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.

[@Khashayarissi-ob4yj]

With luck and more power to you.

[@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.

[@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

[@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

[@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

[@mufafafaexists6782]

an easier solution:
x^3+2x+1=0
by vieta's formula Σα = -b/a = 0
S1 = 0 (where S1 is the sum of roots, S2 is the sum of the squares of the roots S3 is the sum of the cubes of the roots and so on)
by considering each root and summing the 3 results together
α^3+2α+1 = 0
β^3+2β+1 = 0
γ^3+2γ+1 = 0
you get
S3 + 2S1 + 3 = 0
S3 + 2*0 + 3 = 0
S3 = -3
dividing the original equation by x results in
x^2 + 2 + 1/x = 0
rewriting for S
S2 + 6 + S(-1) = 0 {note that S(-1) is the sum of the reciprocals of the roots}
by vieta's again S(-1) = -c/d = -2
solving for S2
S2 + 6 - 2 = 0
S2 = -4
multiplying the original equation by x^2 results in
x^5 + 2x^3 + x^2 = 0
rewriting for S
S5 + 2S3 + S2 = 0
S5 +2*-3 + -4 = 0
S5 = 6 + 4 = 10
therefore α^5 + β^5 + γ^5 = 10

[@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

[@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

[@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

[@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

[@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

[@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 😎