Polynom as sum/product of symmetric polynoms

Question

I have a polynom $(x_1^2x_3 + x_2^2x_1 + x_3^2x_2)(x_1^2x_2 + x_2^2x_3 + x_3^2x_1)$ and I need to express as sum/product of elemental symmetric polynoms $s_1,s_2,s_3$. I know there is an algoritm for that, but can't understand how it works. Can somebody explain it to me?

Answer 1

$(x_1^2x_3 + x_2^2x_1 + x_3^2x_2)(x_1^2x_2 + x_2^2x_3 + x_3^2x_1) = (x_1^4x_2x_3 + x_2^4x_1x_2 + x_3^4x_1x_2) + 3(x_1x_2x_3)^2 + ((x_1x_2)^3 + (x_2x_3)^3+(x_1x_3)^3).$

Let's deal first with $(x_1x_2)^3 + (x_2x_3)^3+(x_1x_3)^3 = A$.

$s_2^2 = (x_1x_2)^2 + (x_2x_3)^2 + (x_1x_3)^2 + 2(x_1^2x_2x_3 + x_1x_2^2x_3 + x_1x_2x_3^2) = (x_1x_2)^2 + (x_2x_3)^2 + (x_1x_3)^2 + 2s_3s_1$

So $(x_1x_2)^2 + (x_2x_3)^2 + (x_1x_3)^2 = s_2^2 - 2s_1s_3$.

Now $s_2(s_2^2 - 2s_1s_3) = (x_1x_2 + x_2x_3 + x_1x_3)((x_1x_2)^2 + (x_2x_3)^2 + (x_1x_3)^2) = (x_1x_2)^3 + (x_2x_3)^3 + (x_1x_3)^3 + x_1x_2^2x_3^3 + x_1x_2^3x_3^2 + x_1^2x_2x_3^3 + x_1^2x_2^3x_3 + x_1^3x_2x_3^2 + x_1^3x_2^2x_3 = A + s_3(x_1x_2^2 + x_1^2x_2 + x_2x_3^2 + x_2^2x_3 + x_1x_3^2 + x_1^2x_3).$

The only thing left is to calculate $x_1x_2^2 + x_1^2x_2 + x_2x_3^2 + x_2^2x_3 + x_1x_3^2 + x_1^2x_3.$

$x_1x_2^2 + x_1^2x_2 + x_2x_3^2 + x_2^2x_3 + x_1x_3^2 + x_1^2x_3 = (x_1x_2^2 + x_1^2x_2) + (x_2x_3^2 + x_2^2x_3) + (x_1x_3^2 + x_1^2x_3) = x_1x_2(x_1 + x_2) + x_2x_3(x_2 + x_3) + x_1x_3(x_1 + x_3) = x_1x_2(s_1-x_3) + x_2x_3(s_1 - x_1) + x_1x_3(s_1 - x_2) = s_1s_2 - 3s_3.$

And finally $s_2(s_2^2 - 2s_1s_3) = A + s_3(s_1s_2 - 3s_3)$, so $A = s_2(s_2^2 - 2s_1s_3) - s_3(s_1s_2 - 3s_3).$

$x_1^4x_2x_3 + x_2^4x_1x_2 + x_3^4x_1x_2 = s_3(x_1^3 + x_2^3 + x_3^3)$ shouldn't be more difficult. Try it yourself!