Polynomial $x^3-2x^2-3x-4=0$ Let $\alpha,\beta,\gamma$ be three distinct roots of the polynomial $x^3-2x^2-3x-4=0$. Then find $$\frac{\alpha^6-\beta^6}{\alpha-\beta}+\frac{\beta^6-\gamma^6}{\beta-\gamma}+\frac{\gamma^6-\alpha^6}{\gamma-\alpha}.$$
I tried to solve with Vieta's theorem. We have 
$$\begin{align}
\alpha+\beta+\gamma &= 2, \\ 
\alpha\beta+\beta\gamma+\gamma\alpha &= -3, \\ 
\alpha\beta\gamma &= 4.
\end{align}$$
For example, $\alpha^2+\beta^2+\gamma^2=(\alpha+\beta+\gamma)^2-2(\alpha\beta+\beta\gamma+\gamma\alpha)=10$ and similarly, we can find $\alpha^3+\beta^3+\gamma^3$... 
But it has very long and messy solution. Can anyone help me?

The final answer I got is 608.
The answer for this question is provided  in the given pdf 
By Vieta's theorem and using some special algebraic identities that are prevelant in India..
Click on the link .THE whole values are achieved by some easy calculations
