Roots of Cubic Equation: Finding $x_1,\,x_2$, and $x_3$

[anemone]

The roots $x_1,\,x_2$ and $x_3$ of the equation $x^3+ax+a=0$ where $a$ is a non-zero real number, satisfy $\dfrac{x_1^2}{x_2}+\dfrac{x_2^2}{x_3}+\dfrac{x_3^2}{x_1}=-8$. Find $x_1,\,x_2$ and $x_3$.

 

[anemone]

Spoiler: Solution of other

We are given the following:
$x_1^3x_3+x_2^3x_1+x_3^3x_2=-8x_1x_2x_3,\\x_1+x_2+x_3=0,\\x_1x_2+x_2x_3+x_3x_1=a,\\x_1x_2x_3=-a$ and for $i=1,\,2,\,3$, $x_i^3+ax_i+a=0$.

Now

$x_1^3+ax_1+a=0\\x_2^3+ax_2+a=0\\x_3^3+ax_3+a=0$ gives

$(x_1^3x_3+x_2^3x_1+x_3^3x_2)+a(x_1x_3+x_2x_1+x_3x_2)+a(x_3+x_2+x_1)=0$

i.e. $8a+a^2=0,\implies a=-8$.

So the given equation is $x^3-8x-8=0$. One root is $-2$ and the other roots are given by $x^2-2x-4=0$, i.e. $x=1\pm \sqrt{5}$.