Finding roots $x_1$, $x_2$, $x_3$ of $x^3+ax+a=0$ (for real $a\neq 0$) if $\sum_{cyc}\frac{x_1^2}{x_2}=-8$

Question

This question is taken from AMTI 1994:

The solutions $x_1$, $x_2$, $x_3$ of the equation $x^3+ax+a=0$, where $a\ne0$ is real, satisfy $$\frac{x_1^2}{x_2}+\frac{x_2^2}{x_3}+\frac{x_3^2}{x_1}=-8$$ Find $x_1$, $x_2$, $x_3$.

I tried to solve this by using Vieta's relations and tried to factorize the the equation they gave, but wasn't able to something solid out of that factorization.

Welcome to the website! Please add your efforts to solve the problem in more detail. Otherwise the question is likely to get downvoted and closed by the community. It is not advised to attach pictures of questions: use Mathjax to format your equations in future (I have done it for you this time). — Shubham Johri, Nov 04 '20 at 11:11
$a=-8$ and roots are $\left(-2,1-\sqrt{5},1+\sqrt{5}\right)$ — Raffaele, Nov 04 '20 at 11:34
Hint: $\sum\limits_{i=1}^3 \frac{x_i^2}{x_{i+1}} = \sum\limits_{i=1}^3 \frac{x_i^3}{x_ix_{i+1}}$ where $x_4 = x_1$. this will allow you to compute $a$. — achille hui, Nov 04 '20 at 12:28

score 0 · Answer 1 · answered Nov 04 '20 at 12:32

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Hint: If $x$ is a root, $x^3=-a(x+1)$ $$\implies -8a = \sum_{cyc} x_1^3x_3 = -a\sum_{cyc} (x_1+1)x_3 = -a\left(\sum_{cyc} x_1x_3 + \sum_{cyc}x_3 \right)=a^2$$

answered Nov 04 '20 at 12:32

Macavity

46,381

−a∑cyc(x1+1)x3 how did u do that ? – AdhirajSinghBrar Nov 05 '20 at 03:49
@AdhirajSinghBrar By replacing $x_1^3$ in the previous expression with $-a(x_1+1)$. – Macavity Nov 05 '20 at 05:24

Finding roots $x_1$, $x_2$, $x_3$ of $x^3+ax+a=0$ (for real $a\neq 0$) if $\sum_{cyc}\frac{x_1^2}{x_2}=-8$

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