Finding the value of $\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha$

Question

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.

Answer 1

Let $\lambda=\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha$ and $\mu=\alpha^2\gamma+\gamma^2\beta+\beta^2\alpha$; then $\lambda+\mu$ and $\lambda\mu$ are symmetric polynomials in $\alpha,\beta,\gamma$ and can be expressed in terms of the elementary symm. pol.-s \begin{align*}s_1&=\alpha+\beta+\gamma&&=-q/p,\\s_2&=\alpha\beta+\beta\gamma+\gamma\alpha&&=\phantom{-}r/p,\\s_3&=\alpha\beta\gamma&&=-s/p\end{align*} (using the approach of a known proof of the fundamental theorem of symm. pol.-s): $$\lambda+\mu=s_1s_2-3s_3,\quad\lambda\mu=s_1^3s_3-6s_1s_2s_3+s_2^3+9s_3^2.$$ This gives a quadratic equation whose roots are $\lambda$ and $\mu$.

Answer 2

If you are very patient and use brute force fot the solution of cubic equation, you should get $$\frac{3 p^2 s-p q r+i \sqrt{p^2 \left(27 p^2 s^2-18 p q r s+4 p r^3+4 q^3 s-q^2 r^2\right)}}{2 p^3}$$ but $$27 p^2 s^2-18 p q r s+4 p r^3+4 q^3 s-q^2 r^2=-p^2\Delta$$ so $$\alpha^2\beta+\beta^2\gamma+\gamma^2\alpha=\frac{3 p^2 s-p q r+i \sqrt{- p^2 \Delta}}{2 p^3}=\frac{3 p s- q r+ \sqrt{ \Delta}}{2 p^2}$$

Answer 3

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 $(2)$.