This involves factoring from Vieta's Formulas:
The first step is to break \(a^3+b^3+c^3\) into \(3r_1r_2r_3+(r_1+r_2+r_3)[r_1^2+r_2^2+r_3^2-(r_1r_2+r_2r_3+r_3r_1)\).
The confusing part might be trying to find \(r_1^2+r_2^2+r_3^2\), yet we know that is equal to \((r_1+r_2+r_3)^2-2(r_1r_2+r_2r_3+r_3r_1).\)
This can be better written as \(3r_1r_2r_3+(r_1+r_2+r_3)[(r_1+r_2+r_3)^2-3(r_1r_2+r_2r_3+r_3r_1)].\)
Remember that \(r_1, r_2\), and \(r_3\) are the roots of the polynomial which is an expression.
Note that \(r_1+r_2+r_3=\frac{-b}{a}\) , \(r_1r_2+r_2r_3+r_3r_1=\frac{c}{a}\), and \(r_1r_2r_3=\frac{-d}{a}.\)
Try to plug the values in, and be careful and don't forget the \(x^2\) term.