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