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?
Best Answer
By long division, the remainder of $x^6$, when divided by $x^3-2x^2-3x-4$, is $77x^2+100x+96$. So we know that $$ \alpha^6=77\alpha^2+100\alpha+96 $$ and the same with other roots. Therefore you are looking at the sum $$ \begin{aligned} S&=\frac{77(\alpha^2-\beta^2)+100(\alpha-\beta)}{\alpha-\beta}+\text{cyclic}\\ &=77(\alpha+\beta)+100+\text{cyclic}\\ &=154(\alpha+\beta+\gamma)+300\\ &=608 \end{aligned} $$ by the (Vieta) relations you have.