$\sum_{i=1}^{i=n}\prod_{j=1,j\neq{i}}^{j=n}{\frac{1}{x_{i}-x_{j}}}=0$
The product on the denominator is what I derive from the derivative of an n degree polynomial with different solutions $x_{1}$ to $x_{n}$. And the sum seems to be zero. I can prove the equation when n is 2 or 3 but I have no idea when n is larger.
When n is 3, I know that $\frac{1}{(x_{1}-x_{2})(x_{1}-x_{3})}$ can break down to $\frac{1}{(x_{1}-x_{2})(x_{2}-x_{3})}-\frac{1}{(x_{1}-x_{3})(x_{2}-x_{3})}$, consequently proving the equation.
I will be very grateful if someone can help me solve this problem or give me some hint.
Assume that the $x_i$ are pairwise different (otherwise the LHS is meaningless). Then, by Lagrange's interpolation formula, $$1=\sum_{i=1}^n\prod_{j\ne i}\frac{x-x_j}{x_i-x_j}.$$ Taking the $x^{n-1}$ coefficient of both sides recovers the desired identity.