Finding a quadratic equation using roots 
If $x_1$ and $x_2$ are the roots of
  $$ax^2+bx+c=0$$ 
  then $x_1^3$ and $x_2^3$ are the roots of which equation?

So I tried by solving this for $x_{1/2}$ so I could change it in $(x-x_1^3)(x-x_2^3)$
$x_{1/2}=\large{-b\pm{\sqrt{4ac}}\over2a}$
and from here:
$$\begin{align}x_1^3&=\bigg({-b+{\sqrt{4ac}}\over2a}\bigg)^3\\&={(\sqrt{4ac}-b)^2(\sqrt{4ac}-b)\over8a^3}\\&={(4ac-2b\sqrt{4ac}+b^2)(\sqrt{4ac}-b)\over8a^3}\\&={4ac\sqrt{4ac}-4abc-8abc-2b^2\sqrt{4ac}+b^2\sqrt{4ac}-b^3\over8a^3}\\&={4ac\sqrt{4ac}-12abc-b^2\sqrt{4ac}-b^3\over8a^3}\end{align}$$
but from here I realized it's probably pointless to do this since I wouldn't be able to use it, and I'm out of ideas.

HINT
We have
$$(x-x_1)(x-x_2)=x^2-(x_1+x_2)x+x_1x_2$$
$$(x-x_1^3)(x-x_2^3)=x^2-(x_1^3+x_2^3)x+x_1^3x_2^3$$
and 
$$x_1^3x_2^3=(x_1x_2)^3$$
$$x_1^3+x_2^3=?$$
