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  "body": "\n## Statement\n\nLet $a, b, c, d \\in \\C$ with $a \\ne 0$.\nPut\n$$p = \\frac{3ac - b^2}{3a^2}, \\qquad q = \\frac{2b^3 - 9abc + 27a^2 d}{27a^3}.$$\nThen $x$ is a [[root|00EHX6FWC0]] of $ax^3 + bx^2 + cx + d$ if and only if $t = x + \\frac{b}{3a}$ is a root of $t^3 + pt + q$.\nIf $p = q = 0$, the only root of $t^3 + pt + q$ is $t = 0$.\nOtherwise, let $\\delta$ be a square root of $\\frac{q^2}{4} + \\frac{p^3}{27}$, chosen so that $w = -\\frac{q}{2} + \\delta \\ne 0$.\nLet $u$ be a cube root of $w$ and let $\\omega = e^{2\\pi i/3}$.\nThen the three roots of $t^3 + pt + q$, counted with multiplicity, are\n$$t_k = \\omega^k u - \\frac{p}{3\\omega^k u}, \\qquad k = 0, 1, 2.$$\n\n## Remarks\n\nA choice of $\\delta$ with $w \\ne 0$ exists: if both choices gave $w = 0$, then $\\delta = 0$ and $q = 0$, hence $p = 0$.\nFor the proof, put $v = -p/(3u)$, so that $uv = -p/3$ and $(u + v)^3 = u^3 + v^3 - p(u + v)$.\nThe 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.\nThe same holds for $\\omega^k u$ in place of $u$, and expanding $\\prod_k (t - t_k)$ gives $t^3 + pt + q$.\nExample: 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 $t_0 = 2 + 1 = 3$.\nWhen the cubic has real coefficients and three distinct real roots, $\\delta$ is not real, and the formula passes through nonreal numbers.\nBy the [[casus irreducibilis|MJ864PBRRJ]], this detour cannot be avoided when the cubic is irreducible over $\\Q$.\n\n## Explanations\n\n### L1 human\n\nThe cubic formula gives the roots of any equation of degree $3$ using arithmetic, square roots and cube roots.\nFirst a shift of $x$ removes the $x^2$ term.\nThen the root is split as a sum $u + v$ of two cube roots, and $u^3$ and $v^3$ solve a quadratic equation.\n\n### L2 human\n\nThe formula shows that every cubic is [[solvable by radicals|J7YS0DA3HQ]], as [[Galois' theorem|VHCWCFA3MX]] predicts, since the Galois group lies in $S_3$, which is solvable.\nThe quantity under the square root is $-\\frac{1}{108}$ times the discriminant of $t^3 + pt + q$, which is $-4p^3 - 27q^2$.\nThe same method of shifting and resolving into a lower-degree equation leads to the [[quartic formula|4CDFMKKZ01]].\n",
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