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There's a fairly well-known formula for cubic equations of the form $x^3+px+q=0$: The transformation $x = z-\frac{p}{3z}$ gives $z^3+q-\frac{p^3}{27}z^{-3}$, and after multiplying by $z^3$, the equation becomes a quadratic in $z^3$. $z$ can be isolated by using the quadratic formula then taking the cube root, and then we can substitute that value into our original transformation to get the roots.

My question is, is there a formula for cubics of the form $x^3+px^2+q=0$ (without a linear term) that can be derived in a similar way, preferably one that doesn't involve eliminating the quadratic term and applying the formula above?

detroitsigma of geometry dash
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    You could obtain such a formula yourself, by replacing $x$ by $1/x$, and then using the formula for cubics without quadratic term. :) I don't know that there is a citeable reference for this. :) – paul garrett Dec 16 '25 at 20:54
  • When $p\ne 0$ and $q\ne 0$, setting $$x=-\frac{\sigma_1\cdot\sigma_2\cdot\sqrt{\sigma_1\cdot 3pq}}{2\cdot p\cdot t}$$ where $\sigma_1^2=\sigma_2^2=1$ and $\sigma_1\cdot 3pq>0$ leads to the equation $$4\cdot t^3+\sigma_1\cdot 3\cdot t=w=\frac{\sigma_2\cdot 3\cdot\sqrt{\sigma_1\cdot 3pq}}{2\cdot p^2}$$ whose real roots can be expressed using identities for $\cos\left(3\cdot\eta\right)$, $\sin\left(3\cdot\eta\right)$, $\cosh\left(3\cdot\eta\right)$, $\sinh\left(3\cdot\eta\right)$ depending on $\sigma_1$ and whether or not $\left|w\right|<1$. – mezzoctane Dec 16 '25 at 21:38
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    You can replace $x$ with $x-\frac{p}3$ to obtain a depressed cubic, or simply go as per @paulgarrett 's way. – Integreek Dec 17 '25 at 05:08

1 Answers1

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Let $y=1/x$, then

$(1/y^3)+(p/y^2)+q=0\implies 1+py+qy^3=0\implies y^3+(p/q)y+(1/q)=0.$

Then use the formula for the "standard" reduced case.

Oscar Lanzi
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