If $a + b + c = 0$, prove that
1)$$ \sum_{\text{cyc}}{\frac{4bc - a^2}{bc + 2a^2}} = 3 $$
2)$$ \prod_{\text{cyc}}{\frac{4bc - a^2}{bc + 2a^2}} = 1 $$
There is a solution that uses two cubic equations. First is of form $x^3+px-q=0$ that has the roots $a,b$ and $c$. After he is forming another cubic equation with the roots the components of the sum of the product. I can't figure out how they've come to the equation $y^3-3y^2-3\frac{5p^3-27q^2}{4p^3+27q^2}y-1=0$
From $\displaystyle x^3+px-q=0, abc=q, ab+bc+ca=p$
and $\displaystyle a^3+pa-q=0\ \ \ \ (1)\iff a^3=q-pa\ \ \ \ (2)$
Assuming $q\ne0,$ $\displaystyle y=\frac{4bc-a^2}{bc+2a^2}=\frac{4abc-a^3}{abc+2a^3}$
$\displaystyle y=\frac{4q-a^3}{q+2a^3}$
Express $a^3$ in terms of $y$ and compare the value of $a^3$ with that of $(2)$ to express $a$ in terms of $y$
Replace the value of $a$ in $(1)$