How to evaluate discriminant of $x^3 +px+q$ given its roots?

Question

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?

Answer 1

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: $$\Delta=q^2+\frac{4p^3}{27}=\frac{27q^2+4p^3}{27}$$

Answer 2

$$f(x)=x^3+qx+r=0 \implies f'(x)=3x^2+q=0 \implies x_{1,2}=\sqrt{-q/3}$$ $$f''(x_1)=6x_1>0 \implies f_{min}=f(x_1), f_{max}=f(x_2)$$ $f(x)=0$ has real roots if $f_{max} >0$ and $f_{min}<0$, then $$f(x_2)f(x_1)0 \implies [-(-q/3)^{3/2}-q\sqrt{-q/3}+r][(-q/3)^{3/2}+q\sqrt{-q/3}+r]<0$$ $$\implies \left(-\frac{2}{3}q\sqrt{-q/3}+r\right) \left(\frac{2}{3}q\sqrt{-q/3}+r\right)<0\implies r^2+\frac{4q^3}{27} \implies27r^2+4q^3<0 $$

Answer 3

Here is a concise derivation. Per $α_1 + α_2 +α_3=0$ and $α_1 α_2 α_3=-q$ $$(α_1 − α_2)^2 = (α_1 + α_2)^2 - 4α_1 α_2=α_3^2+\frac{4q}{α_3} =\frac p{α_3}\left(\frac{3q}p-α_3\right)$$ Similarly $$(α_2 − α_3)^2 = \frac p{α_1}\left(\frac{3q}p-α_1\right),\>\>\>\>\> (α_3 − α_1)^2 = \frac p{α_2}\left(\frac{3q}p-α_2\right)$$ As a result $$\begin{align} D & = (α_1 − α_2)^2 (α_2 − α_3)^2 (α_3 − α_1)^2\\ & = \frac{p^3}{α_1 α_2 α_3} \left(\frac{3q}p-α_3\right)\left(\frac{3q}p-α_1\right)\left(\frac{3q}p-α_2\right) \\ & =- \frac{p^3}q \bigg[\left(\frac{3q}p\right)^3 + p\left(\frac{3q}p\right)+q \bigg] =- 27q^2-4p^3 \\ \end{align}$$