TITLE: Proving a Polynomial Identity
QUESTION [5 upvotes]: Prove that
$$\sum_{i=1}^{n} \dfrac{{r_{i}}^k P(x)}{P'(r_{i})(x-r_{i})} = x^k$$
where $P(x)$ is an $n$ degree polynomial having distinct roots $\{ r_{i} \}_{i=1}^{n}$ and $k$ is an integer, $0 \leq k \leq (n-1)$

I solved it by taking a polynomial $Q(x) = \displaystyle \sum_{i=1}^{n} \dfrac{{r_{i}}^k P(x)}{P'(r_{i})(x-r_{i})} - x^k$ and noting that it has at most $n-1$ degree but $n$ roots i.e, $r_{1}, \ r_{2}, \ \ldots \ , \ r_{n-1}, \ r_{n}$
Thus $Q(x)$ is identically $0$.
I'm seeking other approaches to the problem.
Any help will be appreciated.
Thanks.

REPLY [5 votes]: OK, as zhoraster wrote, the first method is simply recognizing the Lagrange interpolation polynomial
$$
\prod_{j\ne i}^n \frac{z-r_j}{r_i-r_j} = \frac{P(x)}{P'(x)\cdot(x-r_j)}.
$$
So, we are interpolating the points $(r_i,r_i^k)$ by a polynomial and and wonder why do we get $x^k$ as result. :-)

A different approach, using complex analysis (so it will work for complex polynomials only): let $x$ be a fixed complex number, other than $r_1,\ldots,r_n$, and consider the residues of the function
$$
f(z) = \frac{z^k P(x)}{P(z)\cdot (x-z)}.
$$
The degree of this rational function is $k-(n+1)\le-2$, so the sum of its residues is $0$. 
The function $f(z)$ has simple poles at $r_1,\ldots,r_n$ and $x$, its residues are $\dfrac{r_j^kP(x)}{P'(r_j)\cdot (x-r_j)}$ at $z=r_j$, and
$-x^k$ at $z=x$. So,
$$
\sum \mathrm{Res}\,\, f = 
\sum_{j=1}^n \frac{r_j^kP(x)}{P'(r_j)\cdot (x-r_j)} + (-x^k) = 0.
$$