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It seems possible to guess how Tartaglia discovered the solutions of the three cubic equations we called A), B) and C). He gave some hints, which suggest that the way described here is the right one.

First we have to report that Tartaglia thought he discovered what we express in the formula (*a*+*b*)^{3}=*a*^{3}+3*a*^{2}*b*+3*ab*^{2}+*b*^{3} although it had been found already by an Arabian mathematician in the 12th century.

Tartaglia published the first Italian translation of Euclid's *Elements *in 1543. But he had studied the work of the Greek mathematician before this time. He gave lectures about him in 1536 in Venice and knew his work well.

Euclid's Tenth Book treats irrational numbers and combinations of square roots, and distinguished between those forming a sum and those forming a difference (written in modern notation): \[a+\sqrt{b}, \sqrt{a} + b, \sqrt{a} + \sqrt{b},\] called *binomium* in Latin (from the prefix *bi-,* two, and *nomen,* name) and *binomio* in Italian, and \[a-\sqrt{b}, \sqrt{a} - b, \sqrt{a} - \sqrt{b},\] called *residuum* (remainder) or *recisum* (cut off) in Latin and *residuo* or *reciso* in Italian. Today both kinds are called binomials.

The *quadratic* equation has a solution in the form of a binomio \( a+\sqrt{b}\) or of a residuo \( a-\sqrt{b}.\) This suggested that *cubic* equations might also have a solution in the form of a binomio or of a residuo, however, with *cube roots instead of quadratic ones*. Probably Tartaglia tried the different possibilities, and thus found that the residuo \( \sqrt[3]{a} - \sqrt[3]{b}\) and the binomio \( \sqrt[3]{a} + \sqrt[3]{b}\) led to the solutions he sought.

Friedrich Katscher, "How Tartaglia Solved the Cubic Equation - Tartaglia's Solution," *Loci* (August 2011)