If $\alpha, \beta,\gamma$ be the roots of $2x^3+x^2+x+1=0$, show that..........? 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
 A: 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}
A: Hint:
$$-(2x^3+1)^3=(x^2+x)^3$$
$$-8(x^3)^3-12(x^3)^2-6(x^3)-1=(x^3)^2+x^3+3x^3(-(2x^3+1))$$
Replace $x^3$ with $y$ to form a cubic equation in $y$ whose roots are $\alpha^3=a$ etc.
Let $z=\dfrac1a+\dfrac1b+\dfrac1c-\dfrac2c=\dfrac{ab+bc+ca}{abc}-\dfrac2c$
We can find the values of $abc, ab+bc+ca$ from the cubic equation in $y$
$\dfrac2c=?$
As $c$ is a root of the new cubic equation in $y$, replace the value of $c$ to form a cubic equation in $z$
Can you take it from here?
A: Expressing the two complex and conjugate roots as a function of the third root $\gamma$ ,we obtain
$\alpha=-\frac{2\gamma+1}{4}+I\frac{\sqrt{12\gamma^{2}+4\gamma+7}}{4}$
$\beta=\frac{2\gamma+1}{4}-I\frac{\sqrt{12\gamma^{2}+4\gamma+7}}{4}$
therefore
$\Big(\frac{1}{\gamma^{3}}+I\frac{(2\gamma-1)\sqrt{12\gamma^{2}+4\gamma+7}}{(2\gamma^{2}+\gamma+1)^{3}}\Big)$*
$\Big(\frac{1}{\gamma^{3}}-I\frac{(2\gamma-1)\sqrt{12\gamma^{2}+4\gamma+7}}{(2\gamma^{2}+\gamma+1)^{3}}\Big)$*
$\Big(-\frac{1}{\gamma^{3}}+ \frac{(2\gamma+1)(8\gamma^{2}+2\gamma+5)}{(2\gamma^{2}+\gamma+1)^{3}}\Big)=16$
that is
$2\gamma^{3}+\gamma^{2}+\gamma+1=0$.
