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.
 A: 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
$$
(x_1-x_2)^2(x_1-x_3)^2(x_2-x_3)^2 = -(a^3-6a^3+9a^3+27b^2)=-(4a^3+27b^2)
$$
A: 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. 
A: 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
$$(x_1-x_2)^2(x_2-x_3)^2(x_3-x_1)^2=-4a^3-27b^2$$
