Proof of an Algebraic Identity

Question

Let $x_1\dots x_n$ be a set of nonequal real numbers with $n>1$. Does the following algebraic identity hold? $$0=\sum_{i=1}^n\prod_{\substack{j=1\\j\neq i}}^{n}1/(x_i-x_j)$$ It seems to work for the first few cases. If so, does anyone have a simple proof? Beyond noticing that this is symmetric in the $x_i$, I can't come up with anything.

Answer 1

Consider the polynomial $-1+\sum_{i=1}^n\prod_{\substack{j=1\\j\neq i}}^{n}(x-x_j)/(x_i-x_j)$.

It has degree at most $n-1$. Substituting any of $x_1,x_2,\dots,x_n$ evaluates to 0. A nonzero degree at most $n-1$ polynomial has at most $n-1$ roots, so it must be identically 0. Looking at the leading coefficient gives the desired result.