How Tartaglia Solved the Cubic Equation - Cubic Equations

Editor’s note: This article was published in December 2006.

Today we have one cubic equation, which we represent

\(x^3+nx^2+px+q=0\) (\(n\), \(p\), \(q\) positive, negative, or zero),

and which comprises all possible cases.  But imagine living at a time when \(0\) was only a digit, and not regarded as a number, and, therefore, setting equal to zero was unknown.

And imagine moreover that negative numbers, and also negative solutions of equations, were rejected—called false or fictitious—because at that time one thought geometrically, and the side of a square, or the edge of a cube cannot be negative. It was the French-Dutch mathematician Albert Girard (1595–1632), who in his 1629 book Invention nouvelle en algebre (New invention in algebra) was the first to explain the minus sign geometrically: "The solution with minus is explained in geometry by retrograding, & the minus goes back, where the plus advances." He also urged, "the solutions with minus should not be omitted." But the consequence of geometric thinking in the 16th century was that, with rare exceptions, only positive terms, only plus signs, were allowed on both sides of an equation.

That meant that, instead of the single cubic equation of today, there were at least 13 equations, 7 with all four terms (cubic, quadratic, linear, and absolute term), 3 without the linear term, and 3 without the quadratic term:

7 complete cubic equations (all powers represented):
\(x^3+nx^2+px=q\)
\(x^3+nx^2+q=px\)
\(x^3+px+q=nx^2\)
\(x^3+nx^2=px+q\)
\(x^3+px=nx^2+q\)
\(x^3+q=nx^2+px\)
\(x^3=nx^2+px+q\)

3 equations without the linear term:
\(x^3+nx^2=q\)
\(x^3=nx^2+q\)
\(x^3+q=nx^2\)

3 equations without the quadratic term:
A) \(x^3+px=q\)
B) \(x^3=px+q\)
C) \(x^3+q=px\)

Each type of cubic equation was treated separately. But, in the early 1500s, the ten cubic equations containing the quadratic term were too difficult to be solved. At first, only the three types we call here A), B) and C) were accessible.