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?
 A: 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,
$$\begin{align}
D 
& = (α_1 − α_2)^2 (α_2 − α_3)^2 (α_3 − α_1)^2\\
& = \frac{p^3}{α_1 α_2 α_3} \left(\frac{3q}p-α_3\right)\left(\frac{3q}p-α_1\right)\left(\frac{3q}p-α_2\right) \\
& =- \frac{p^3}q  \left(\left(\frac{3q}p\right)^3 + p\left(\frac{3q}p\right)+q \right) \\
& =- 27q^2-4p^3 \\
\end{align}$$
A: This is simply Cardano's method that you can find here: https://en.m.wikipedia.org/wiki/Cubic_equation. 
In particular, you have to set $x=u+v$, obtainig:
$$(u+v)^3+p(u+v)+q=0$$
Picking up $3uv$, we have:
$$u^3+v^3+3uv(u+v)+p(u+v)+q=0$$
And so:
$$u^3+v^3+(3uv+p)(u+v)+q=0$$
Now, we impose this condition:
$$\left\{\begin{matrix}
u^3+v^3=q
\\ 3uv=-p
\end{matrix}\right.$$
which leads to a quadratic equation of the form $t^2-qt+\frac{p^3}{27}=0$ where $t=v^3$ with: $$\Delta=q^2+\frac{4p^3}{27}=\frac{27q^2+4p^3}{27}$$
A: $$f(x)=x^3+qx+r=0 \implies f'(x)=3x^2+q=0 \implies x_{1,2}=\sqrt{-q/3}$$
$$f''(x_1)=6x_1>0 \implies f_{min}=f(x_1), f_{max}=f(x_2)$$
$f(x)=0$ has real roots if $f_{max} >0$ and $f_{min}<0$, then
$$f(x_2)f(x_1)0 \implies [-(-q/3)^{3/2}-q\sqrt{-q/3}+r][(-q/3)^{3/2}+q\sqrt{-q/3}+r]<0$$
$$\implies \left(-\frac{2}{3}q\sqrt{-q/3}+r\right) \left(\frac{2}{3}q\sqrt{-q/3}+r\right)<0\implies r^2+\frac{4q^3}{27} \implies27r^2+4q^3<0 $$
