The story of Bombelli solving the equation $x^3=15x+4$ in L'Algebra and introducing $\sqrt{-1}$ is well known. The equation, having obvious root $x=4$, is solved using Cardano's formula: $$ x=\sqrt[3]{\frac{q}{2}+\sqrt{(\frac{q}{2})^2-(\frac{p}{3})^3}}+\sqrt[3]{\frac{q}{2}-\sqrt{(\frac{q}{2})^2-(\frac{p}{3})^3}} $$ getting $x=\sqrt[3]{2+11\sqrt{-1}}+\sqrt[3]{2-11\sqrt{-1}}$.

After that, all the accounts I've read so far merely check that $$ (2+\sqrt{-1})^3=2+11\sqrt{-1} $$ and $$(2-\sqrt{-1})^3=2-11\sqrt{-1}$$ to conclude that the formula gives the correct answer $x=(2+\sqrt{-1})+(2-\sqrt{-1})=4$.

The question is: how did Bombelli transform $\sqrt[3]{2\pm11\sqrt{-1}}$ into $2\pm\sqrt{-1}$? Is it detailed in the book? Did he know a procedure to simplify $\sqrt[3]{m+n\sqrt{k}}$ in the form $a+b\sqrt{k}$ for positive $k$? Was this procedure exposed in earlier chapters of L'Algebra, or in other books of the time, in relation to Cardano's formula?

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    $\begingroup$ I believe that you meant to say $x^3=15x+4$ rather than $x^3+15x=4$. The short article Bombelli and the Invention of Complex Numbers is devoted to the methods Bombelli used to transform one expression into the other. $\endgroup$
    – nwr
    Sep 21, 2019 at 2:13
  • $\begingroup$ You're right of course, I fixed the equation. I will look into the article. $\endgroup$
    – Adrien
    Sep 21, 2019 at 4:01

1 Answer 1


Bombelli did not have a general method for such transformation, nor does it exist. Doing so is equivalent to solving the irreducible case of the cubic, which is impossible in (real) radicals. It is solvable in inverse trigonometric functions, as Vieta showed later, but simplifying his trigonometric expression is itself a chore.

However, when everything is an integer one can find the solutions by educated guessing. In this case, according to Katz's History of Mathematics, 12.3.2, Bombelli used indeterminates $a,b$ to set up (in modern notation) $(a+\sqrt{-b})^3=2+11\sqrt{-1}$. Given his rules for $\sqrt{-1}$, this quickly leads to $(a-\sqrt{-b})^3=2-11\sqrt{-1}$, and then to $$a^2+b=5;\ \ a(a^2-3b)=2.$$ By simple divisibility considerations, there are only four choices for $a$, and Bombelli shows that $a=2$ is the only viable one, with $b=1$. Note that attempting to solve this "honestly" leads to another cubic $4a^3=15a+2$, no better than the original one.

LaNave-Mazur's Reading Bombelli gives more insight into the way Bombelli handled cubics, and complex numbers that came up in them. As they note, Bombelli's "procedure" was generally not very tidy:

"if you open Bombelli's treatise you discover nothing resembling complex numbers until page 133, at which point certain mathematical objects (that might be regarded by a modern as "complex numbers") burst onto the scene, in hill battle array, in the middle of an on-going discussion. Here is how Bombelli introduces these mathematical objects. He writes, "I have found another sort of cubic radical which behaves in a very different way from the others"... Complex numbers, when they occur in Gerolamo Cardano's earlier treatise Ars Magna, occur neatly as quantities like $2 + \sqrt{-15}$. But they appear initially in Bombelli's treatise as cubic radicals of the type of quantities discussed by Cardano; a somewhat complicated way for them to arise in a treatise that is thought of as an organized exposition of the formal properties of complex numbers!"


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