If $\alpha$, $\beta$, $\gamma$ are roots of $x^3+p x+q=0$, then $\alpha^5+\beta^5+\gamma^5=5\alpha\beta\gamma(\alpha\beta+\beta\gamma+\gamma\alpha)$.

Question

If $\alpha$, $\beta$, $\gamma$ are the roots of $x^3+px+q=0$, show that $$\alpha^5+\beta^5+\gamma^5=5\alpha\beta\gamma\ (\alpha\beta+\beta\gamma+\gamma\alpha)$$

I tried to solve using Vieta's formula, but I was getting a sign error.

From the equation $x^3+px+q=0$, $$\begin{align} \sum \alpha\phantom{\beta} &= \phantom{-}0 \\ \sum \alpha\beta &= \phantom{-}p \\ \prod α \phantom{\beta} &= -q \end{align}$$

L.H.S. $$\begin{align} \alpha^5+\beta^5+\gamma^5&=(\alpha^4+\beta^4+\gamma^4)(\alpha+\beta+\gamma)-\alpha^4(\beta+\gamma)-\beta^4(\gamma+\alpha)-\gamma^4(\alpha+\beta) \\ \text{(with $\sum \alpha = 0$)}\quad &=-\alpha\beta\gamma\left(\begin{array}{c} \phantom{+}\left( \dfrac{\alpha^3}{\beta} +\dfrac{\alpha^3}{\gamma} +\dfrac{\beta^3}{\gamma} +\dfrac{\beta^3}{\alpha} +\dfrac{\gamma^3}{\alpha} +\dfrac{\gamma^3}{\beta}\right)\\ +\left( \dfrac{\alpha^3}{\alpha} -\dfrac{\alpha^3}{\alpha} +\dfrac{\beta^3}{\beta} -\dfrac{\beta^3}{\beta} +\dfrac{\gamma^3}{\gamma} -\dfrac{\gamma^3}{\gamma}\right)\end{array}\right) \\ &=-\alpha\beta\gamma\left(\left(\alpha^3+\beta^3+\gamma^3\right) \left(\dfrac {1}{\alpha}+\dfrac {1}{\beta}+\dfrac {1}{\gamma}\right) -\left(\alpha^2+\beta^2+\gamma^2\right)\right) \tag{1} \end{align}$$

Putting $\sum \alpha = 0$ in the formula $$\begin{align} \alpha^3+\beta^3+\gamma^3-3\alpha\beta\gamma &=\left(\alpha+\beta+\gamma\right)\left(\alpha^2+\beta^2+\gamma^2-\alpha\beta-\beta\gamma-\gamma\alpha\right) \\ \alpha^3+\beta^3+\gamma^3&=3\alpha\beta\gamma \tag{2} \end{align}$$

Putting (2) in (1) $$\begin{align} &\phantom{=}\;-\prod\alpha \cdot\left(3\cdot \prod \alpha \cdot \dfrac{\sum\alpha\beta}{\prod\alpha}-\left(\left(\sum\alpha\right)^2-2\cdot\sum\alpha\beta\right)\right) \\ &=-\prod\alpha\cdot\left(3\cdot\sum\alpha\beta+2\cdot\sum\alpha\beta\right) \\ &=-5\cdot\sum\alpha\beta\cdot \prod\alpha \end{align}$$

There is a negative sign in my solution. Please point out where I have gone wrong, or share your solution.

Answer 1

plugging $$\gamma=-\alpha-\beta$$ in the left-hand sode we get $$-5\,\alpha\,\beta\, \left( \beta+\alpha \right) \left( {\alpha}^{2}+ \alpha\,\beta+{\beta}^{2} \right) $$ (after factirization) and the right-hand side: $$5\,\alpha\,\beta\, \left( \beta+\alpha \right) \left( {\alpha}^{2}+ \alpha\,\beta+{\beta}^{2} \right) $$