<s>[INST] 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.
 [/INST] Let
$$f(x)=x^3+ax+b=(x-x_1)(x-x_2)(x-x_3)$$
so that
$$f'(x)=3x^2+a=(x-x_1)(x-x_2)+(x-x_1)(x-x_3)+(x-x_2)(x-x_3)$$
This allows us to get the equation
$$ f'(x_1)f'(x_2)f'(x_3)=(3x_1^2+a)(3x_2^2+a)(3x_3^2+a)=-(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2 $$
We try to evaluate
$$ 
\begin{align}
(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2 &=
-(3x_1^2+a)(3x_2^2+a)(3x_3+a) \\
&=-(a^3+3a^2(x_1^2+x_2^2+x_3^2)+9(x_1^2 x_2^2+x_1^2 x_3^2+x_2^2x_3^2)a+27x_1^2x_2^2x_3^2) 
\end{align}
$$
Using the Vieta's formulas
$$
\begin{align}
x_1+x_2+x_3 &= 0\\
x_1x_2+x_1x_3+x_2x_3 &= a\\
x_1x_2x_3 &=-b
\end{align}
$$
we obtain
$$
\begin{align}
x_1^2+x_2^2+x_3^2 &= (x_1+x_2+x_3)^2 - 2(x_1x_2+x_1x_3+x_2x_3)\\
 &=-2a\\
x_1^2x_2^2+x_1^2x_3^2+x_2^2x_3^2 &= (x_1x_2+x_1x_3+x_2x_3)^2-2x_1x_2x_3(x_1+x_2+x_3)\\
&= a^2
\end{align}
$$
Hence we get the final expression
$$
(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2 = -(a^3-6a^3+9a^3+27b^2)=-(4a^3+27b^2)
$$
</s>