Cubic formula

Remarks

A choice of $\delta$ with $w \ne 0$ exists: if both choices gave $w = 0$, then $\delta = 0$ and $q = 0$, hence $p = 0$. For the proof, put $v = -p/(3u)$, so that $uv = -p/3$ and $(u + v)^3 = u^3 + v^3 - p(u + v)$. The numbers $u^3 = w$ and $v^3 = -p^3/(27w)$ are the two roots of $z^2 + qz - p^3/27$, so $u^3 + v^3 = -q$ and $u + v$ is a root. The same holds for $\omega^k u$ in place of $u$, and expanding $\prod_k (t - t_k)$ gives $t^3 + pt + q$. Example: for $t^3 - 6t - 9$, the number $\frac{q^2}{4} + \frac{p^3}{27} = \frac{81}{4} - 8 = \frac{49}{4}$; with $\delta = \frac{7}{2}$, $w = 8$, $u = 2$ and $t_0 = 2 + 1 = 3$. When the cubic has real coefficients and three distinct real roots, $\delta$ is not real, and the formula passes through nonreal numbers. By the casus irreducibilis, this detour cannot be avoided when the cubic is irreducible over $\Q$.

Going deeper

For a student

The formula shows that every cubic is solvable by radicals, as Galois' theorem predicts, since the Galois group lies in $S_3$, which is solvable. The quantity under the square root is $-\frac{1}{108}$ times the discriminant of $t^3 + pt + q$, which is $-4p^3 - 27q^2$. The same method of shifting and resolving into a lower-degree equation leads to the quartic formula.