Use Viete's relations to prove the roots of the equation $x^3+ax+b=0$ satisfy $(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2=-4a^3-27b^2$ Use Viete's relations to prove that the roots $x_1$, $x_2$, and $x_3$ of the equation $x^3+ax+b=0$ satisfy the identity $(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2=-4a^3-27b^2$.
I know that viete's relations state that the roots $x_1$, $x_2$, and $x_3$ of the equation $x^3-px^2+qx-r=0$ have the property $p=x_1+x_2+x_3$, $q=x_1x_2+x_1x_3+x_2x_3$ and $r=x_1x_2x_3$.
My question is whether or not there is a way to do this without multiplying out $(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2$ and showing that it factors into $4(x_1+x_2+x_3)^3+27(x_1x_2x_3)$ because that algebra involved in that looks like it will be nasty.

If $b=0$, then the roots are $0,\pm\sqrt{-a}$ from which
 $$(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2=-4a^3$$
follows.
In the following, $b\not=0$. 
We have, by Vieta's formulas,
$$x_1+x_2+x_3=0,\quad x_1x_2+x_2x_3+x_3x_1=a,\quad x_1x_2x_3=-b$$
So, we can have $$\begin{align}x_3(x_1-x_2)^2&=x_3((x_1+x_2)^2-4x_1x_2)\\\\&=x_3(-x_3)^2-4x_1x_2x_3\\\\&=x_3^3-4(-b)\\\\&=(-ax_3-b)-4(-b)\\\\&=-ax_3+3b\end{align}$$
Similarly, we get
$$x_1(x_2-x_3)^2=-ax_1+3b,\qquad x_2(x_3-x_1)^2=-ax_2+3b$$
It follows from these that
$$\begin{align}&x_1x_2x_3(x_1-x_2)^2(x_2-x_3)^2(x_3-x_1)^2\\\\&=x_3(x_1-x_2)^2x_1(x_2-x_3)^2x_2(x_3-x_1)^2\\\\&=(-ax_3+3b)(-ax_1+3b)(-ax_2+3b)\\\\&=-a^3x_1x_2x_3+3a^2b(x_1x_2+x_2x_3+x_3x_1)-9ab^2(x_1+x_2+x_3)+27b^3\\\\&=-a^3(-b)+3a^2b\cdot a-9ab^2\cdot 0+27b^3\end{align}$$
Dividing the both sides by $x_1x_2x_3=-b\not=0$ gives
$$(x_1-x_2)^2(x_2-x_3)^2(x_3-x_1)^2=-4a^3-27b^2$$
