Summation of Powers over Product of Differences/Example

Proof
Instance of Summation of Powers over Product of Differences:
 * $\displaystyle \sum_{j \mathop = 1}^n \paren {\dfrac { {x_j}^r} {\displaystyle \prod_{\substack {1 \mathop \le k \mathop \le n \\ k \mathop \ne j} } \paren {x_j - x_k} } } = \begin{cases}0 & : 0 \le r < n - 1 \\ 1 & : r = n - 1 \\ \displaystyle \sum_{j \mathop = 1}^n x_j & : r = n \end{cases}$

where:
 * $n = 3$
 * $x_1 = a$
 * $x_2 = b$
 * $x_3 = c$