Express a given symmetric polynomial as a polynomial in elementary symmetric polynomials

Let n \geq 3, and let x_i be indeterminates for 1 \leq i \leq n. Express p = \sum_{i=1}^n x_i^3 as a polynomial in the elementary symmetric functions \sigma_1 = \sum_{i=1}^n x_i, \sigma_2 = \sum_{i < j} x_ix_j, and \sigma_3 = \sum_{i < j < k} x_ix_jx_k.


We claim that p = \sigma_1^3 - 3\sigma_1\sigma_2 + 3\sigma_3.

Consider \sigma_1\sigma_2. We have the following:

\sigma_1\sigma_2  =  \left( \sum_{i=1}^n x_i \right) \left( \sum_{j < k} x_jx_k \right)
 =  \sum_{j < k} \sum_{i=1}^n x_ix_jx_k
 =  \sum_{j < k} \left( x_j^2x_k + x_jx_k^2 + \sum_{i \neq j,k} x_ix_jx_k\right)
 =  \sum_{j < k} x_j^2x_k + \sum_{j < k} x_jx_k^2 + \sum_{j < k} \sum_{i \neq j,k} x_ix_jx_k
 =  \sum_{i \neq j} x_i^2x_j + 3 \sum_{i < j < k} x_ix_jx_k
 =  \sum_{i \neq j} x_i^2x_j + 3 \sigma_3.

For brevity, let q = \sum_{i \neq j} x_i^2x_j; thus \sigma_1\sigma_2 = q + 3\sigma_3. Moreover, we see that

\sigma_1^3  =  \left( \sum_{i=1}^n x_i \right)^3
 =  \sum_{i,j,k} x_ix_jx_k
 =  \sum_{i=1}^n x_i^3 + 3 \sum_{i \neq j} x_i^2x_j + 6\sum_{i < j < k} x_ix_jx_k
 =  p + 3q + 6\sigma_3.

So we have p = \sigma_1^3 - 3q - 6\sigma_3 and q = \sigma_1\sigma_2 - 3\sigma_3; a quick substitution reveals that p = \sigma_1^3 - 3\sigma_1\sigma_2 + 3\sigma_3.

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