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Before we examine the solution to \(x^3=ax^2+b,\) we consider some background on the cubic.

In 1494, Luca Pacioli published his *Somma di aritmetica, geometrica, proporzioni e proporzionalità, *which collected much of the mathematical knowledge of the time. In his account of algebra, he listed the cubic problems, including those unsolved at the time. He also put forth his opinion that the unsolved problems would never be solved. Note that I write “cubic problems” rather than “cubic equations”: equations, as we know them, were a product of the hundred years *after* Cardano wrote, culminating with Descartes and analytic geometry. And I use the plural, “cubic *problems*,” because at that time, the theory of negative numbers had not been fleshed out (Cardano called them *ficta* or “fictitious” in places). Thus, the cubic problems were all stated with strictly positive coefficients, and there was never a question of moving all terms to one side to achieve any kind of standard form. What made these problems “unsolved” was the lack of a formula for the solution in terms of the numbers given in the problem; geometric solutions for some of the cubic problems had been constructed by Islamic mathematicians, such as Umar al-Khayyami’s solution of 1070 [Berggren, p. 119].

Niccolo Tartaglia (1500-1557) (Source: |

The rather dramatic story of the solution of the cubic has been told, excellently, elsewhere. (See [Dunham, pp. 133-142] or [Katz, pp. 358-361], for example.) In brief, Scipione del Ferro, a professor at Bologna, found the solution to "cube and thing equal to number" (in our symbolic notation, \(x^3=ax+b\) with \(a\) and \(b\) positive rational numbers) in the early sixteenth century. He told his student Antonio Maria Fior of the solution, and Fior challenged Niccolo Tartaglia to a mathematical duel. In such a duel, each contestant posed a list of questions to the other, with the loser paying the winner a prize. Tartaglia had found the solution to "cube and square equal to number," and, just in time, the solution to "cube equal to thing and number." Fior couldn’t answer Tartaglia's questions, and Tartaglia won the duel. A few years later, Cardano obtained the solution to both cubics from Tartaglia (whether or not such obtaining was underhanded is still a matter of contention), and published the solutions in his *Ars Magna*.

The solution I will present is to the problem “cube equal to square and number,” or \(x^3=ax^2+b.\) Cardano’s solution in the *Ars Magna* proceeds in three steps over three chapters. First, in Chapter 6, Cardano decomposed the cube into eight parts, establishing a geometric demonstration for our binomial formula \[{(x+y)}^3=x^3+3x^2y+3xy^2+y^3.\]He used this decomposition in almost all of his arguments about cubics. Second, in Chapter 14, Cardano depressed the problem “cube equal to square and number” to “cube equal to thing and number”: symbolically, he depressed \(x^3=ax^2+b\) to \(y^3=Ay+B.\) His geometric procedure was radically different from our analytic one, and relied on the decomposition of the cube from Chapter 6. This depression allowed him to use his solution to “cube equal to thing and number” presented in Chapter 12, to derive his rule for the solution to “cube equal to square and number.”

His solution in *Ars Magna* to "cube equal to thing and number" consists of two parts, a geometric demonstration that a particular number is the solution, and a rule for finding that number from the coefficients of the problem. The demonstration does not show how to find the number given by the rule; I argue that the number arises naturally from an *abbaco* "problem of ten," and that Cardano relied on this *abbaco* knowledge to connect his demonstration with his rule.

William B. Branson (St. Cloud State University), "Solving the Cubic with Cardano - The Problem," *Loci* (September 2013)