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?