TITLE: About the roots of cubic polynomial
QUESTION [8 upvotes]: Let $\alpha, \beta, \gamma$ be the complex roots of the polynomial $P_3(x)=ax^3+bx^2+cx+d$. 
Is there any known formula for calculating
$\alpha^2 \beta+\beta^2 \gamma+ \gamma^2\alpha \; , \;
\alpha \beta^2+\beta \gamma^2+\gamma\alpha^2$
(in terms of $a,b,c,d$)?
If no, can someone obtain it?

REPLY [22 votes]: Let $u=\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha$ and $v=\alpha\beta^2+\beta\gamma^2+\gamma\alpha^2$. We have the following relations between the elementary symmetric polynomials $e_1$, $e_2$ and $e_3$ of $\alpha$, $\beta$ and $\gamma$ and the coefficients of $P_3$:
$$
e_1 = \alpha+\beta+\gamma = -\frac{b}{a} \\
e_2 = \alpha\beta+\beta\gamma+\gamma\alpha = +\frac{c}{a} \\
e_3 = \alpha\beta\gamma = -\frac{d}{a}
$$
We find
\begin{eqnarray*}
p & :=  u+v = & e_1e_2-3e_3 \\
q & :=  uv = & e_1^3e_3-6e_1e_2e_3+e_2^3+9e_3^2
\end{eqnarray*}
From this, you can obtain $u$ and $v$ solving $t^2-pt+q=0$.