Product of cubic root differences

Question

Given a cubic equation $x^3 + ax^2 + bx + c$, with three real roots $r_1, r_2, r_3$, how can we express $(r_1 - r_2)(r_2 - r_3)(r_3 - r_1)$ in terms of $a,b,c$ and how do we prove it using Vieta's? Assume $r_1 \leq r_2 \leq r_3$

Answer 1

Let $r_1+r_2+r_3=3u=-a$, $r_1r_2+r_1r_3+r_2r_3=3v^2=b,$ where $v^2$ can be negative, and $r_1r_2r_3=w^3=-c$.

Thus, $$(r_1-r_2)^2(r_1-r_3)^2(r_2-r_3)^2=27(3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6).$$ Can you end it now?

Answer 2

If you square that, you get the discriminant of the cubic which is $a^2b^2+18abc-4b^3-4a^3c-27c^2$.

It would be worth proving that this is indeed true using vieta's formulae though.