<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's first prove Vieta's cubic relations.
$\underline {\text{Proof}}$: Let the roots of a cubic polynomial, $f(x)$, be $\alpha$, $\beta$, $\gamma$. Then $f(x) = (x-\alpha)(x-\beta)(x-\gamma)$. Let $f(x) = x^3 - px^2 + qx - r$.
\begin{align}
f(x) & = (x-\alpha)(x-\beta)(x-\gamma) \\
& = (x^2 - \beta x - \alpha x + \alpha \beta) (x - \gamma) \\
& = (x^3 - \gamma x^2 - \beta x^2 + \beta \gamma x - \alpha x^2 + \alpha \gamma x + \alpha \beta x - \alpha \beta \gamma) \\
& = x^3 - (\alpha + \beta + \gamma)x^2 + (\alpha \beta + \alpha \gamma + \beta \gamma)x - (\alpha \beta \gamma) \\
\implies p & = \alpha + \beta + \gamma \\
q & = \alpha \beta + \alpha \gamma + \beta \gamma \\
r & = \alpha \beta \gamma
\end{align}

The proof of a cubic discriminant is quite long and there is no easy way to do it, so I'll link a proof here.
Now with the question.

$\underline {\text{Proof}}$: A cubic polynomial $f(x) = (x-\alpha)(x-\beta)(x-\gamma) = x^3 - px^2 + qx - r$ has a discriminant 
\begin{align}
\Delta_3 & = (\alpha-\beta)^2 (\alpha-\gamma)^2 (\beta-\gamma)^2 \\
\end{align}
Assume $p=0$. We have that $q$ and $r$ are polynomials of degrees $2$ and $3$. The discriminant is a polynomial of degree $6$ (look at the definition just above), and hence the discriminant must be a linear combination of the one-term polynomials, $q^3$ and $r^2$. 
$$\Delta_3 = mq^3 + nr^2$$
where $m$ and $n$ are two constants. Now, we can do something clever to get our final result. Let $q = -1$ and $r=0$. Now we have a new polynomial
\begin{align}
0 & = x^3 - x \\
& = x(x^2 - 1)\\
& = x(x-1)(x+1)\\
\end{align}
with roots $-1, 0, 1$.
\begin{align}
\Delta_3 & = (\alpha-\beta)^2 (\alpha-\gamma)^2 (\beta-\gamma)^2 \\
& = (-1 - 0)^2 (-1 - 1)^2 (0 - 1)^2\\
& = 1 \cdot 4 \cdot 1\\
& = 4 \\
\end{align}
So far, if you've been keeping up, we have $\Delta = 4q^3 + br^2$.
In a similar method, if we set $q=0$, $r=−1$ we get the polynomial $x^3−1=0$. Now we are in the territory of complex numbers! Solving for roots of $x^3-1=0$ we get $1, \omega, \omega^2$. $\omega$ is a third root of unity.
\begin{align}
\Delta_3 & = (\alpha-\beta)^2 (\alpha-\gamma)^2 (\beta-\gamma)^2 \\
& = (1 - \omega)^2 (1 - \omega^2)^2 (\omega - \omega^2)^2 \\
\end{align}
Finishing this we get the discriminant is equal to $27$. 
Hence, we have 
\begin{align}
\Delta_3 &= 4q^3 + 27r^2 \\
& = 4(\alpha \beta + \alpha \gamma + \beta \gamma)^3 + 27(\alpha \beta \gamma)^2 \\
\end{align}
And we are done. 
</s>