If $\alpha, \beta,\gamma$ be the roots of $2x^3+x^2+x+1=0$, show that:
$$(\frac{1}{\beta^3}+\frac{1}{\gamma^3}-\frac{1}{\alpha^3})(\frac{1}{\gamma^3}+\frac{1}{\alpha^3}-\frac{1}{\beta^3})(\frac{1}{\alpha^3}+\frac{1}{\beta^3}-\frac{1}{\gamma^3})=16$$
Here's what I have tried,
By Vieta's rule
$\alpha+\beta+\gamma=\frac{-1}{2}\text{. ………..}(1)$
$\alpha \beta+\beta \gamma+\alpha \gamma=\frac{1}{2}\text{. ………..}(2)$
$\alpha \beta \gamma=\frac{-1}{2}\text{. ………..}(3)$
Squaring $(1)$,
$\alpha^2+\beta^2+\gamma^2+2(\alpha\beta+\beta\gamma+\alpha\gamma)=\frac{1}{4}\text{. ………..}(4)$
From $(2)$,
$\alpha^2+\beta^2+\gamma^2=\frac{-3}{4}\text{. ………..}(5)$
Putting the roots and adding these equations,
$2\alpha^3+\alpha^2+\alpha+1=0$
$2\beta^3+\beta^2+\beta+1=0$
$2\gamma^3+\gamma^2+\gamma+1=0$
We get,
$2(\alpha^3+\beta^3+\gamma^3)+(\alpha^2+\beta^2+\gamma^2)+(\alpha+\beta+\gamma)+3=0$
Putting the values,
$2(\alpha^3+\beta^3+\gamma^3)+\frac{-3}{4}+\frac{-1}{2}+3=0$
$(\alpha^3+\beta^3+\gamma^3)=\frac{-7}{8}$
Then I divided $2x^3+x^2+x+1=0$ by $x$ and I found out $(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma})$ the same way I found out $(\alpha^3+\beta^3+\gamma^3)$. Likewise doing the same, finding $\sum\frac{1}{\alpha^2}$ then $\sum\frac{1}{\alpha^3}$ I found out
$\sum\frac{1}{\alpha^3}=-4$
But still I'm far from the answer, also the $-$ sign is creating problems.
Any help would be highly appreciated
Best Answer
Notice that if we multiplly the polynomial with $x-1$ we get $$x^3(x-1)+(x-1)(x^3+x^2+x+1)=0$$ so $$2x^4-x^3-1=0\implies \boxed{{1\over x^3}= 2x-1}$$
So your expression is $$E=(2\beta + 2\gamma -2\alpha -1)(2\beta -2\gamma +2\alpha -1)(-2\beta + 2\gamma +2\alpha -1)$$
Now notice that $$\beta + \gamma +\alpha =-{1\over 2}$$ so \begin{align}E &= -8(2\alpha +1)(2\beta +1)(2\gamma +1)\\ &=-64 (\alpha +{1\over 2})(\beta +{1\over 2})(\gamma +{1\over 2}) \\ &=-32\cdot p(-{1\over 2})=-16\end{align}