## 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 $p = \sigma_1^3 - 3\sigma_1\sigma_2 + 3\sigma_3$.