Hard cubic equation

Question

If $$x=a+b+c;y=-(ab+bc+ac); z=-abc$$ solve the following equation: $$t^3-xt^2-yt+z=0$$ I was thinking to use the identity $$a^3+b^3+c^3-3abc=\frac{1}{2}[(a-b)^2+(b-c)^2+(a-c)^2]$$ but I don't know how. Maybe it's not useful but $x$, $y$ and $z$ are solutions of the following system $$a^2x+ay-z=a^3$$ $$b^2x+by-z=a^3$$ $$c^2x+cy-z=c^3$$

Answer 1

Since\begin{align}(t-a)(t-b)(t-c)&=t^3-a t^2-b t^2-c t^2+a b t+a c t+b c t-a b c\\&=t^3-xt^2-yt+z,\end{align}the roots are $a$, $b$, and $c$.