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

Xander Henderson
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Ederson
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4 Answers4

6

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 $\operatorname{tr} A^5 = 5$.

copper.hat
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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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$z_1^3=z_1+1,z_2^3=z_2+1$

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

$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 $

ole
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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.

Micah Windsor
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  • Anyone care to explain the downvote? – Micah Windsor Apr 09 '20 at 17:41
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    Cardano is probably the worst method to use here. Think about raising the roots to the fifth power ! –  Apr 09 '20 at 17:44
  • @YvesDaoust You can raise any number to the fifth power using a calculator with ease. What are you on about? – Micah Windsor Apr 09 '20 at 17:45
  • If a CAS or a numerical solver is allowed, there is no point doing the exercise. There is a solution without computing the roots. –  Apr 10 '20 at 09:46