Decompose the symmetric polynomial $\Sigma{x_1^2x_2^2x_3^2}$ into elementary symmetric polynomials. The method I am attempting to use involves either (when not all exponents are equal, for example $\Sigma{x_1x_2^2}$) repeatedly extracting the monomial with highest possible equal exponents (so for the unequal exponent example above $(\Sigma{x_1})(\Sigma{x_1x_2})$) or moving the exponent outside the summation when all exponents are equal, as in the question title, i.e. the one I'm asking about, so the first step here is $\Sigma{x_1^2x_2^2x_3^2}$ = $(\Sigma{x_1x_2x_3})^2$.
Obviously this is $E_3^2$, together with the terms that need to be subtracted off, based on how many variables are in common between the 2 $E_3$'s in $E_3^2$: 0, 1, or 2.
If none are in common, you can use either of the $E_3$'s to determine the choice of 3 out of the total of 6 indeterminates, so that term is $2E_6$.
My thinking was, if 1 indeterminate is in common you get an expression that needs to be further broken down, i.e. $\Sigma{x_1^2x_2x_3}$, that will be multiplied by $E_2$.
Similarly, if If 2 indeterminates are in common you get an expression that needs to be further broken down, i.e. $\Sigma{x_1^2x_2^2x_3}$, that will be multiplied by $E_1$.
Thus far my attempt to solve this seems to be maybe heading towards the book answer, which is $E_3^2 + 2E_1E_5 - 2E_2E_4 -2E_6$.
But my next step in further decomposing $\Sigma{x_1^2x_2x_3}$ and $\Sigma{x_1^2x_2^2x_3}$ led to much more complicated terms, with no cancellation that I could see to lead to the simpler set of terms involving just $E_1E_5$, $E_2E_4$, and $E_6$ to subtract off of $E_3^2$. Also, the book is adding the $E_1E_5$ term back, suggesting that there is a sequence of decompositions I'm getting wrong, maybe involving cancellation.
Can anyone show where I'm getting this wrong?
 A: The key to your error is that each set of $E_6$ does not simply come up only twice, it actually comes up ${6 \choose 3} = 20$ times. On the other hand, given a $\sum x_1^2x_2x_3x_4x_5$ there are actually ${4 \choose 2} = 6$ ways to set up the same expression, while $\sum x_1^2x_2^2x_3x_4$ has only two ways to be set up. Also, the new monomials created are not as simple as your examples, which is easy to see since the entire expression should be degree 6.
To explain, given a monomial $abcdef$ in $E_6$, you can create this monomial through $abc \cdot def$, $abd \cdot cef$, etc. All ways of choosing 3 elements from 6 works. Given $abcde^2$ in $E_5E_1$, you can create the monomial through $abe \cdot cde$, $ace \cdot bde$, etc. All ways of choosing 2 elements from 4 works. This exact process is used to determine the coefficients in the computation below.
Since this computation is so error-prone, I will just do the entire calculation from start to finish, then you can check your results against these steps.
Notation: $S_n = E_n, P_{a, b, c, ...} = \sum \limits_{\text{sym}} x_1^ax_2^bx_3^c...$, where $S_n$ is an alternative notation for elementary symmetric polynomials and $P_{a,b,c...}$ is the Muirhead-type abbreviation.
$P_{2,2,2} = S_3^2 -2P_{2,2,1,1} - 6P_{2,1,1,1,1} - 20S_6$ (Result 1)
$P_{2,2,1,1} = S_2S_4-4P_{2,1,1,1,1}-15S_6$ (Result 2)
$P_{2,1,1,1,1} = S_1S_5-6S_6$ (Result 3)
$P_{2,2,2} = S_3^2 -2S_2S_4 + 2P_{2,1,1,1,1} +10S_6$ (Using results 1 and 2 -> result 4)
$P_{2,2,2} = S_3^2 -2S_2S_4 + 2S_1S_5 - 2S_6$ (Using results 4 and 3 -> Answer)
And we are done. All it is is careful work and computation, nothing crazy.
