Scipione del Ferro (ca. 1465 – 1526) is the first person known to have solved cubic equations algebraically. However, he did not disclose this remarkable achievement until he was near death. The momentous solution fell to the ears of Antonio Maria Fior, a student of del Ferro, who, upon gaining the information, began to boast that he was able to solve cubic equations. This delusion of grandeur gave rise to a public problem-solving contest between Fior and Niccolo Fontana or Tartaglia as he was more widely known. It was at this contest, set in Venice on the 22 February 1535 that Fior posed problem to Tartaglia.
There was a tree of height 12, which was broken in two in such a way that the height of the part remaining was the cube root of the part cut away. What was the height of the part remaining? Fortunately for the latter, a moment of inspiration had come ten days previously. Tartaglia had discovered a method of solving two types of cubic, those of the form a cube and things equal to numbers () and a cube and squares equal to numbers (). Fior could only solve equations in the first form (del Ferro’s type) and so the solution of other type set by his opponent eluded him. Tartaglia, armed with more tools reduced his challenges into special cases of the cubic and quickly solved all thirty of Fior’s problems in less than two hours.
In 1548 Tartaglia became involved in another public contest, this time his opponent was Ludovico Ferrari. It was to be held in Milan on the 10th August at the church garden of the Fratti Zoccolanti in front of a large, distinguished audience. It was here that the following problem was posed. Divide 8 into two parts such that their product multiplied by their difference comes to as much as possible.
There was a tree of height 12, which was broken in two in such a way that the height of the part remaining was the cube root of the part cut away. What was the height of the part remaining? This problem translates to the cubic equation A formula, developed by Cardan can be used to solve a cubic in this form. His formula states that for a cubic equation in the form However, this method will only give one value for x. In order to solve for all values of x another formula must be employed. Firstly it is necessary to determine the nature of the roots of the quadratic. Equation (?) has one real root and a pair of complex roots and these roots can be determined by the formulas where each of the variables are determined by a, b, c and d correspond to real numbers in the general quadratic.
Fortunately equation (?) is a little simpler than the above form because it has no x2 term, this implies that b = 0. If the number 12 is taken away from both sides to give d=-12 then, together with the previous two, the numbers a = 1 and c = 1 can be substituted into the above formulas to obtain f = 1, g = -12, h = , Now these numbers can be substituted into equations (?) and the following values are obtained for the roots of the cubic polynomial: These results are likely to be quite different from the ones originally given. Firstly Tartaglia would only have looked for one value x and only found one value of x. In fact roots of negative numbers never occurred to Tartaglia and were not understood by Cardan. Tartaglia would also have used a different method of solution and would not have given his solution in the above algebraic form.