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Let real numbers x_1, x_2, \cdots , x_n satisfy 0 < x_1 < x_2 < \cdots< x_n < 1 and set x_0 = 0, x_{n+1} = 1. Suppose that these numbers satisfy the following system of equations:
\sum_{j=0, j \neq i}^{n+1} \frac{1}{x_i-x_j}=0 \quad \text{where } i = 1, 2, . . ., n.
Prove that x_{n+1-i} = 1- x_i for i = 1, 2, . . . , n.

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