Evaluate discriminant of $x^3 +px+q$ given its roots? Consider the cubic polynomial $x^3 +px+q$ with $p, q ∈ \mathbb{Q}$. Suppose $α_1, α_2, α_3$ are the roots and thus $$x^3 + px + q = (x − α_1)(x − α_2)(x − α_3).$$
Let $D = (α_1 − α_2)^2(α_1 − α_3)^2(α_2 − α_3)^2$.
(a) Express $p$ and $q$ in terms of $α_1, α_2$ and $α_3$.
(b) Show that $D = −4p^3 − 27q^2$
I have that $p= α_1α_2 + α_1α_3 + α_2α_3 $ and $ q= -α_1α_2α_3$
Is there a more succinct way to prove part (b) without expanding out a ridiculously large polynomial by brute force?

Per $α_1 + α_2 +α_3=0$ and $α_1 α_2 α_3=-q$,
$$(α_1 − α_2)^2 = (α_1 + α_2)^2 - 4α_1 α_2=α_3^2+\frac{4q}{α_3}
=\frac p{α_3}\left(\frac{3q}p-α_3\right)$$
Likewise, $(α_2 − α_3)^2 = \frac p{α_1}\left(\frac{3q}p-α_1\right)$ and $
(α_3 − α_1)^2 = \frac p{α_2}\left(\frac{3q}p-α_2\right)$. Then,
$$\begin{align}
D 
& = (α_1 − α_2)^2 (α_2 − α_3)^2 (α_3 − α_1)^2\\
& = \frac{p^3}{α_1 α_2 α_3} \left(\frac{3q}p-α_3\right)\left(\frac{3q}p-α_1\right)\left(\frac{3q}p-α_2\right) \\
& =- \frac{p^3}q  \left(\left(\frac{3q}p\right)^3 + p\left(\frac{3q}p\right)+q \right) \\
& =- 27q^2-4p^3 \\
\end{align}$$
