If $x^3 +px -q =0$ then value of $(\alpha + \beta)(\beta + \gamma)(\alpha+\gamma)(1/\alpha^2 + 1/\beta^2+1/\gamma^2)$ I am given a cubic equation $E_1 : x^3 +px -q =0$ where $p,q \in R$ so what would be value of the expression  $$(\alpha + \beta)(\beta + \gamma)(\alpha+\gamma)(\frac{1}{\alpha^2} + \frac{1}{\beta^2}+\frac{1}{\gamma^2})$$ where $\alpha , \beta,\gamma$ are roots of cubic equation $E_1$
I know that this has got to do something with sum/product of root of equation but I don't know how to solve this problem , perhaps I can do something by assuming value of $p,q$ to be something then solve the given expression?
 A: Hint:Observe that : $\alpha^3 + \beta^3 + \gamma^3 = 3q - p(\alpha + \beta + \gamma) = 3q-p(0) = 3q$, and $(\alpha+\beta)(\beta + \gamma)(\gamma + \alpha) = \dfrac{(\alpha+\beta+\gamma)^3- (\alpha^3+\beta^3+\gamma^3)}{3} = \dfrac{0^3 - p^2-3q}{3}$, and $\alpha\beta\gamma = -q$, and also $\alpha\beta+\beta\gamma+\gamma\alpha = p$
A: This one is easy. You need to note that
\begin{align}
\alpha + \beta + \gamma &= 0\tag{1}\\
\alpha\beta + \beta\gamma + \gamma\alpha &= p\tag{2}\\
\alpha\beta\gamma &= q\tag{3}
\end{align}
and hence the calculation of the desired expression $E$ is given by
\begin{align}
E &= (\alpha + \beta)(\beta + \gamma)(\gamma + \alpha)\left(\frac{1}{\alpha^{2}} + \frac{1}{\beta^{2}} + \frac{1}{\gamma^{2}}\right)\\
&= (-\gamma)(-\alpha)(-\beta)\left(\frac{1}{\alpha^{2}} + \frac{1}{\beta^{2}} + \frac{1}{\gamma^{2}}\right)\\
&= -\frac{\alpha^{2}\beta^{2} + \beta^{2}\gamma^{2} + \gamma^{2}\alpha^{2}}{\alpha\beta\gamma}\\
&= -\frac{(\alpha\beta + \beta\gamma + \gamma\alpha)^{2} - 2\alpha\beta\gamma(\alpha + \beta + \gamma)}{\alpha\beta\gamma}\\
&= -\frac{p^{2}}{q}
\end{align}
A: hint: do you see that $ \alpha \beta +\beta \gamma+ \gamma \alpha = p $ and $\alpha \beta \gamma = q$?
