How to use these to represent the $(x_{1}-x_{2})(x_{2}-x_{3})(x_{3}-x_{1})$

Question

Is there any identity with $x_{1}x_{2}x_{3}$, $x_{1}x_{2}+x_{2}x_{3}+x_{3}x_{1}$, $x_{1}+x_{2}+x_{3}$ and $(x_{1}-x_{2})(x_{2}-x_{3})(x_{3}-x_{1})$? Moreover, if I know the $x_{1}x_{2}x_{3}$, $x_{1}x_{2}+x_{2}x_{3}+x_{3}x_{1}$, $x_{1}+x_{2}+x_{3}$, how to use these to represent the $$(x_{1}-x_{2})(x_{2}-x_{3})(x_{3}-x_{1})$$

Answer 1

$\Delta = (x_1-x_2)(x_2-x_3)(x_3-x_1)$ changes sign when when you swap two of the $x_i$, whereas $s_1 = x_1+x_2+x_3$, $s_2=x_1x_2+x_2x_3+x_3x_1$, $s_3=x_1x_2x_3$ are all unchanged, so there is no way to represent $\Delta$ itself in terms of the $s_i$.

On the other hand, $\Delta^2$ is symmetric in the $x_i$, so the Fundamental Theorem of Symmetric Polynomials says it can be written as a polynomial in the $s_i$. Indeed, one finds $$ \Delta^2 = s_1^2 s_2^2 - 4 s_2^3 - 4 s_1^3 s_3 + 18 s_1 s_2 s_3 - 27 s_3^2 $$ (it has to be some linear combination of terms that have $6$ $x_i$s in them, but then you need to find the coefficients. It's easier to shift the $x_i$ to make $s_1=0$, do the calculation for these, which contains far fewer terms, then shift back and find the general case by expansion.

Answer 2

Since the polynomials $x_1x_2x_3$, $x_1x_2+x_2x_3+x_3x_1$, and $x_1+x_2+x_3$ are symmetric but $(x_1-x_2)(x_2-x_3)(x_3-x_1)$ isn't, you can't express this last polynomial as a function of the other three.

However, $\bigl((x_1-x_2)(x_2-x_3)(x_3-x_1)\bigr)^2$ is symmetric and it is in fact equal to\begin{multline}(x_1+x_2+x_3)^2(x_1x_2+x_2x_3+x_3x_1)^2-4x_1x_2x_3(x_1+x_2+x_3)^3+\\-4(x_1x_2+x_2x_3+x_3x_1)^3+18x_1x_2x_3(x_1x_2+x_2x_3+x_3x_1)(x_1+x_2+x_3)-27(x_1x_2x_3)^2.\end{multline}