TITLE: Finding the value of $\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha$
QUESTION [6 upvotes]: If $\alpha,\beta,\gamma$ are the roots of the cubic polynomial $px^3+qx^2+rx+s$, then how can I find the value of $\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha$ in terms of p,q,r and s?

My attempt:
Using Vieta's formulas,
$$\alpha+\beta+\gamma=\frac{-q}{p}$$
$$\alpha\beta+\beta\gamma+\gamma\alpha=\frac{r}{p}$$
$$\alpha\beta\gamma=\frac{-s}{p}$$
I first thought that this would help. We can find $\alpha\beta\gamma$ by Vieta's formula but the expression inside this bracket could not be possibly found.
$$\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha=\alpha\beta\gamma\left(\frac{\alpha}{\gamma}+\frac{\beta}{\alpha}+\frac{\gamma}{\beta}\right)$$
Then, I found
$$\alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma\alpha)=\frac{q^2-2rp}{p^2}$$
But here all $\alpha^2$, $\beta^2$ and $\gamma^2$ are all multiplied with different factors, so possibility of taking a common factor is ruled out.
Then, I thought to take an example. By chance, I could get a pattern.
For example, take the polynomial $x^3-1$. The roots of this polynomial are $1,\omega,\omega^2$.
Then, in this case the expression evaluates to $1^2\cdot \omega+\omega^2\cdot 1+\omega^4\cdot 1=\omega-1$. The expression in this case gives a complex result, so possibly couldn't be expressed in terms of $p,q,r,s$.
Any hint is appreciated.

REPLY [4 votes]: Consider the polynomial $x^3+Qx^2+Rx+S$ with roots $\alpha,\beta,\gamma$. We have $$(\alpha+\beta+\gamma)(\alpha\beta+\beta\gamma+\gamma\alpha)=(\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha)+3\alpha\beta\gamma+(\alpha^2\gamma+\beta^2\alpha+\gamma^2\beta)$$ so that $$(\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha)+(\alpha^2\gamma+\beta^2\alpha+\gamma^2\beta)=3S-QR.\tag1$$ Further, \begin{align}\small(\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha)(\alpha^2\gamma+\beta^2\alpha+\gamma^2\beta)&=\small\alpha^3\beta^3+\alpha^3\gamma^3+\beta^3\gamma^3+3\alpha^2\beta^2\gamma^2+\alpha\beta\gamma(\alpha^3+\beta^3+\gamma^3)\\&=\small-S^3\left(\frac1{\alpha^3}+\frac1{\beta^3}+\frac1{\gamma^3}\right)+3S^2-S(\alpha^3+\beta^3+\gamma^3).\end{align} We can evaluate the last term by noting that $$(\alpha+\beta+\gamma)^3=\alpha^3+\beta^3+\gamma^3+3(\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha+\alpha^2\gamma+\beta^2\alpha+\gamma^2\beta)+6\alpha\beta\gamma$$ so that $$\alpha^3+\beta^3+\gamma^3=-Q^3-3S+3QR.$$ Using the second of Vieta’s formulae $\alpha\beta+\beta\gamma+\gamma\alpha=R$, dividing both sides by $\alpha\beta\gamma$ yields $1/\alpha+1/\beta+1/\gamma=-R/S$. Since $$\small\left(\frac1\alpha+\frac1\beta+\frac1\gamma\right)^3=\frac1{\alpha^3}+\frac1{\beta^3}+\frac1{\gamma^3}+3\left(\frac1{\alpha^2\beta}+\frac1{\beta^2\gamma}+\frac1{\gamma^2\alpha}+\frac1{\alpha^2\gamma}+\frac1{\beta^2\alpha}+\frac1{\gamma^2\beta}\right)+\frac6{\alpha\beta\gamma},$$ we obtain $$-\frac{R^3}{S^3}=\frac1{\alpha^3}+\frac1{\beta^3}+\frac1{\gamma^3}+3\frac{3S-QR}{S^2}-\frac6S$$ using $(1)$. Thus $$(\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha)(\alpha^2\gamma+\beta^2\alpha+\gamma^2\beta)=R^3+9S^2-6QRS+Q^3S.\tag2$$ We can finally determine the values of both $\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha$ and $\alpha^2\gamma+\beta^2\alpha+\gamma^2\beta$ by simultaneously solving $(1)$ and $(2)$.