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Hello everyone I have the polynomial $z^3 -z -1 = 0$ and $z_1 , z_2 , z_3$ are the roots of this polynomial.

How can I find $z_1^5 + z_2^5 + z_3^5 ?$

I know that $z_1 + z_2 + z_3 = 0 , z_1 \cdot z_2 + z_1 \cdot z_3 + z_2 \cdot z_3 = -1 , z_1z_2z_3 = 1$

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  • $\begingroup$ Have you tried Cardano's method for solving a depressed cubic? $\endgroup$ – Micah Windsor Apr 9 at 17:31
  • $\begingroup$ No. What is it? $\endgroup$ – Yehonatan Apr 9 at 17:33
  • $\begingroup$ I will post an answer. It is quite simple. $\endgroup$ – Micah Windsor Apr 9 at 17:36
  • $\begingroup$ Thank you Micha! $\endgroup$ – Yehonatan Apr 9 at 17:37
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Here is a simple slightly tedious method:

Let $A=\begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix}$ and note that $\chi_A(z) = z^3-z-1$. Hence the sum of the $5$th powers of the roots is given by $\operatorname{tr} A^5 = 5$.

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    $\begingroup$ Rabbits out of a hat. $\endgroup$ – Yves Daoust Apr 9 at 17:35
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    $\begingroup$ That matrix is my companion :-). $\endgroup$ – copper.hat Apr 9 at 17:36
  • $\begingroup$ I didn't learn matrix but thanks. $\endgroup$ – Yehonatan Apr 9 at 17:41
  • $\begingroup$ @YvesDaoust's method is nice. $\endgroup$ – copper.hat Apr 9 at 17:42
  • $\begingroup$ Excellent (+1). $\endgroup$ – Gibbs Apr 9 at 17:45
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Hint:

For all roots, $$z^5=z^2z^3=z^3+z^2=z^2+z+1.$$

Expand $(z_1+z_2+z_3)^2.$

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  • $\begingroup$ @Yehonatan: what is $z^3$ ? $\endgroup$ – Yves Daoust Apr 9 at 17:37
  • $\begingroup$ I didn't tell you i tell cooper $\endgroup$ – Yehonatan Apr 9 at 17:38
  • $\begingroup$ And what do you mean at what is $z^3$ ? $\endgroup$ – Yehonatan Apr 9 at 17:39
  • $\begingroup$ @Yehonatan: observe your problem. $\endgroup$ – Yves Daoust Apr 9 at 17:39
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    $\begingroup$ +1: Nice observation. $\endgroup$ – copper.hat Apr 9 at 17:40
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$z_1^3=z_1+1,z_2^3=z_2+1$

$z_1z_2z_3 = 1\rightarrow z_1z_2(z_1+z_2)=-1$

$z_1^5 + z_2^5 + z_3^5=z_1^5 + z_2^5 - (z_1+z_2)^5= -(5z_1^4z_2+10z_1^3z_2^2+10z_1^2z_2^3+5z_1z_2^4)=-5z_1z_2(z_1^3+2z_1^2z_2+2z_1z_2^2+z_2^3)=-5z_1z_2(z_1+z_2+2+2z_1z_2(z_1+z_2))=-5z_1z_2(z_1+z_2+2-2)=-5z_1z_2(z_1+z_2)= 5 $

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Your equation is of the form: $$y^3+Ay=B$$ where: $$y=z,A=-1,B=1$$ Now substitute: $$3st=A$$ $$s^3-t^3=B$$ Then solve for: $$y=s-t$$ Can you find the roots now? Once you have these roots, you can perform whatever operations on them you like.

P.S. You can find more information about Cardano's method here.

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  • $\begingroup$ Anyone care to explain the downvote? $\endgroup$ – Micah Windsor Apr 9 at 17:41
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    $\begingroup$ Cardano is probably the worst method to use here. Think about raising the roots to the fifth power ! $\endgroup$ – Yves Daoust Apr 9 at 17:44
  • $\begingroup$ @YvesDaoust You can raise any number to the fifth power using a calculator with ease. What are you on about? $\endgroup$ – Micah Windsor Apr 9 at 17:45
  • $\begingroup$ If a CAS or a numerical solver is allowed, there is no point doing the exercise. There is a solution without computing the roots. $\endgroup$ – Yves Daoust Apr 10 at 9:46

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