$x=\sqrt[3]{q+\sqrt{q^2-p^3}}-\sqrt[3]{-q+\sqrt{q^2-p^3}}$
Why is it that this must be true $q^2-p^3<0$ for x to have 3 distinct real roots?
$x=\sqrt[3]{q+\sqrt{q^2-p^3}}-\sqrt[3]{-q+\sqrt{q^2-p^3}}$
Why is it that this must be true $q^2-p^3<0$ for x to have 3 distinct real roots?
This is the casus irreducibilis of the cubic. You need to use complex numbers even if the roots are all real. The proof needs Galois theory.