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I was trying to decompose into elementary symmetric polynomials: $$(x_1^2 - x_2 - x_3 + 1)(x_2^2 - x_1 - x_3 + 1)(x_3^2 - x_1 - x_2 + 1)$$ But it didn't work out for me. I tried to add one of the variables ($x_3$) so that there are fewer terms $$(x_1^2 - x_2 + 1)(x_2^2 - x_1 + 1)(- x_1 - x_2 + 1) = (-s_1 + 1)(s_2^2 + s_2 + 1 + x_1^2 + x_2^2 - x_1^3 - x_2^3 - (x_1 + x_2)) = $$ $$= (-s_1 + 1)(s_2^2 + s_2 + 1 + s_1^2 - 2s_2 - s_1^3 + 3s_1^2s_2 - s_1)$$ But I do not know what to do next. You can try to subtract from the original what I got and further decompose the resulting polynomial into elementary symmetric ones. And the answer will be the sum of the two decompositions obtained. But it seems that it is very long and easy to make mistakes. Tell me, please, what am I doing wrong?

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  • $\begingroup$ All of those good old 19th century math students knew how to do this! $\endgroup$
    – GEdgar
    Commented May 19 at 12:15
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    $\begingroup$ View it as $\prod (x_1^2 + x_1 + 1 - s_1) $, and expand out in powers of $s_1$.EG (The easy case is ) Coefficient of $s_1^2$ is $\sum ( x_1^2 + x_1 + 1)$, which you can easily convert to $s_1^2 - 2s_2 + s_1 + 3$ $\endgroup$
    – Calvin Lin
    Commented May 19 at 13:08

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Fleshing out my comment, though I leave it to you to check the arithmetic. Nothing really insightful in the algebraic expansion.

$\prod ( x_1^2 + x_1 + 1 - s_1) $

$ = -s_1 ^3 + s_1 ^2 [ \sum ( x_1^2 + x_1 + 1)] - s_1 [ \sum ( x_1^2+x_1+1)(x_2^2+x_2+1)] + \prod ( x_1^2 + x_1 + 1 ) $
-> Expanding as a polynomial in $s_1$

$ = - s_1^2 + s_1 [ s_1^2 - 2s_2 + s_1 + 3] - s_1 [ s_2^2 - 2s_3s_1 + s_2s_1 - 3s_3 + 2s_1^2 - 3s_2 + 2s_1 + 3 ] + [s_3^2 + s_3 s_2 + ( s_2)^2 - s_3s_1 + s_2s_1 - 2s_3 - s_1^2 + s_2 + s_1 + 1 ] $
-> Rewrite each coefficient as an elementary symmetric polynomial

$ = -s_1^3 + s_1^2 ( -s_2 + 2s_3 - 3) + s_1( -s_2^2 + 2s_2 + 2s_3) + s_1 + (s_2^2 + s_3^2 + s_2 + s_2s_3 - 2s_3 + 1)$
-> Simplifying as a polynomial in $s_1$

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  • $\begingroup$ I like "heck the arithmetic" even if you meant "check the arithmetic". "Noting insightful" instead of "nothing insightful" too. $\endgroup$ Commented May 19 at 14:18
  • $\begingroup$ and how do you decompose coefficients into elementary symmetric ones so quickly? I can directly multiply brackets and group them, which is not an effective method. $\endgroup$ Commented May 19 at 15:42
  • $\begingroup$ The coefficients for degrees 3 and 2 are easily calculated, and they are also easy to express in terms of elementary symmetric ones, but for others it is not very clear how to do this $\endgroup$ Commented May 19 at 15:50
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    $\begingroup$ @NickSchemov I multiplied then out and then slowly removed the ESO starting from the largest degree. EG If it was a $a^3$ term, then you need $s_1^3$. If it was a $a^2b$ term, then you need $s_2 s_1$. $\quad$ Practice and familiarity also helps, but otherwise there isn't much more insight into the calculation. $\endgroup$
    – Calvin Lin
    Commented May 19 at 21:09

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