Suppose that \mathbf{ x_1 , \cdots , x_n} (n> 2) are real numbers such that x\mathbf{x_i = -x_{n-i+1}} for \mathbf{1\le i \le n} . Consider the sum \mathbf{ S = \sum \sum \sum x_i x_j x_k } where the summations are taken over all i, j, k: \mathbf{ 1\le i, j, k \le n } and i, j, k are all distinct. Then S equals:

(A) \mathbf{n!x_1 x_2 \cdots x_m } ; (B) (n-3)(n-4); (C) (n-3)(n-4)(n-5); (D) none of the foregoing expressions;

Discussion:

\mathbf {( x_1 + x_2 + ... + x_n )^3}

\mathbf{= \sum \sum \sum {x_i x_j x_k }+ \sum x_i ^2 ( \sum x_j ) + \sum x_i^3}

Since \mathbf {x_1 = - x_n}

Hence \mathbf {x_1 ^3 = -x_n ^3}

Since \mathbf{\sum {x_i} = 0 } and \mathbf{\sum {x_i}^3 = 0}

Therefore \mathbf{\sum \sum \sum {x_i x_j x_k } = 0}

Hence option D