Finding the roots of $x^3+ax+a=0$ If the roots $x_1,x_2,x_3$ of the equation $x^3+ax+a=0$ satisfy $$\frac{x_1^2}{x_2}+\frac{x_2^2}{x_3}+\frac{x_3^2}{x_1}=-8$$ then the roots of the equation are?
I wrote the Vieta formulas down but then I'm stuck.Please give your suggestions.

$\frac{x_1^2}{x_2}+\frac{x_2^2}{x_3}+\frac{x_3^2}{x_1}=-8 \iff \frac{x_1^3}{x_1x_2}+\frac{x_2^3}{x_2x_3}+\frac{x_3^3}{x_1x_2}=-8 \iff \frac{-ax_1-a}{x_1x_2}+\frac{-ax_2-a}{x_2x_3}+\frac{-ax_3-a}{x_1x_3}=-8 \iff \frac{a}{x_1x_2}+\frac{a}{x_2x_3} + \frac{a}{x_1x_3} + \frac{a}{x_1}+ \frac{a}{x_2}+ \frac{a}{x_3}=8 (1)$
Because $x_1x_2x_3=-a$ (Vieta), from (1) we have:
$-(x_1 + x_2 +x_3) - (x_1x_2 + x_1x_3+x_2x_3) = 8$
Should be easy from here.
