<s>[INST] Reducing $\frac{a^n(b-c) - b^n(a-c) + c^n(a-b)}{(a-b)(a-c)(b-c)}$ for $n>2$ 
Simplify
$$\frac{a^n(b-c) - b^n(a-c) + c^n(a-b)}{(a-b)(a-c)(b-c)}$$ for $n>2$.

The answer is $(a+b+c)^n$, but I can't seen to get it.
Can someone help me?
Thanks
 [/INST] Hint
$$a-c=(a-b)+(b-c)$$
so
\begin{align*}&(b-c)\cdot a^n-(a-b)\cdot b^n-(b-c)\cdot b^n+(a-b)\cdot c^n\\
&=(b-c)(a^n-b^n)+(a-b)(c^n-b^n)\\
&=(a-b)(b-c)(a^{n-1}+a^{n-2}b+\cdots +b^{n-1}-c^{n-1}-c^{n-2}b-\cdots-b^{n-1})\\
&=(a-b)(b-c)[(a^{n-1}-c^{n-1})+(a^{n-2}b-c^{n-2}b)+\cdots+(ab^{n-2}-cb^{n-2})]\\
&=(a-b)(b-c)(a-c)[(a^{n-2}+\cdots+c^{n-2})+b(a^{n-3}+\cdots+c^{n-3})+\cdots+b^{n-2}]
\end{align*}
</s>