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
 A: 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*}
A: $$\frac{(b-c)\times a^n - (a-c)\times b^n + (a-b)\times c^n}{(a-b)(a-c)(b-c)}=$$
$$\frac{a^n-b^n+c^n}{1}=$$
$$\boxed{\frac{a^n}{(a - b) (a - c)} - \frac{b^n}{(a - b) (b - c)} + \frac{c^n}{(a - c) (b - c)}}.$$
A: The answer is neither $(a+b+c)^n$ nor $(a-b+c)^n$.
As can be found here and  here , we have that
$$
\frac{(b-c)\times a^n - (a-c)\times b^n + (a-b)\times c^n}{(a-b)(a-c)(b-c)}\\
 = \frac{a^n}{(a-b)(a-c)}+\frac{b^n}{(b-a)(b-c)}+\frac{c^n}{(c-a)(c-b)} = 
\displaystyle\sum_{\substack{n_1, n_2, n_3 \geqslant 0 \\ n_1 + n_2 + n_3 = n - 2}}a^{n_1}b^{n_2}c^{n_3}
$$
Some examples:
$$P_2 = 1 \\
P_3 = a + b  +c \\
P_4 = a^2 + b^2 + c^2 + ab + bc + ac 
$$
A: One can think of the given formula as a particular case of the determinant formula for the Schur polynomials, namely
$$
\frac{a^n(b-c) - b^n(a-c) + c^n(a-b)}{(a-b)(a-c)(b-c)}=\frac{\begin{vmatrix}a^n & b^n & c^n \\ a & b & c\\ 1 & 1 & 1\end{vmatrix}}{\begin{vmatrix}a^2 & b^2 & c^2 \\ a & b & c\\ 1 & 1 & 1\end{vmatrix}}=s_{(n-1)}(a,b,c)
$$
It is know that $s_{(n-2)}=h_{n-2}$, where $h_k$ is the complete homogeneous symmetric polynomial of degree $k$ (see here, for example). Thus,
$$
\frac{a^n(b-c) - b^n(a-c) + c^n(a-b)}{(a-b)(a-c)(b-c)}=h_{n-2}(a,b,c)=\sum_{\substack{k,l,m\ge0 \\ k+l+m=n-2}}a^kb^lc^m.
$$
For more information about the Schur polynomials, see, e.g. here.
I realize that this might be too complicated for this problem but I wanted to emphasise that there is a "big theory" behind problems like this.
