# Find the coefficients in $(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2 = s_1^2s_2^2 + as_1^3s_3 + bs_1s_2s_3 + cs_2^3+ds_3^2$

Use evaluation homomorphisms $F[x_1,x_2, \dots, x_n] \to F$ to obtain the coefficients in: $$(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2 = s_1^2s_2^2 + as_1^3s_3 + bs_1s_2s_3 + cs_2^3+ds_3^2$$ where the $s_i$ are symmetric polynomials.

I'm getting really frustrated with this problem. Since I'm working in any field, all I can plug in is the values $0, 1, -1$. If more than one of $x_1, x_2, x_3$ are zero, then the equation reduces to $0=0$. I have tried the combinations $0,0,1; ~ 0,1,-1; ~ 0,-1,-1; ~ 1,1,1; ~ 1,1,-1; ~ 1,-1,-1; ~ -1,-1,-1$ and get the equations relating the coefficients $c = -4, -5 = -a+b+d$ and $27 = 27a + 9b +d$. Clearly, I need one more, but I can't find any other combination of numbers to plug in to obtain a new one. What have I missed?

• @user26857 It seems I was using $s_2^3 in my calculations, it was just a typo here. Do you have any other idea what could be wrong? – Johanna Feb 21 '15 at 2:32 • You don't need to limit yourself to$0, 1, -1$. It is clear that if you solve your problem for$F = \mathbb Z$(nevermind that this is not a field -- the theorem that every symmetric polynomial is a polynomial in the elementaries works over any commutative ring), then the coefficients$a,b,c,d$you get will work over every$F$. And you can solve linear systems better over$\mathbb Z$. – darij grinberg Feb 21 '15 at 2:36 • @darijgrinberg Thank you! I solved it..$a=-4, b = 18, c = -4, d = -27\$. – Johanna Feb 21 '15 at 3:05