# How Tartaglia Solved the Cubic Equation - Tartaglia's Solution in Modern Notation

Author(s):
Friedrich Katscher

We describe first, in modern notation, the solution method for equation A) x3+px=q: If you cube the residuo, the difference of two cube roots,  $\sqrt[3]{u} - \sqrt[3]{v},$ where at that time always $u > v,$ you get ${\big( {\sqrt[3]{u} - \sqrt[3]{v}}\big)}^3 = u - 3{\sqrt[3]{u}}{\sqrt[3]{v}}\big( {\sqrt[3]{u}}-{\sqrt[3]{v}}\big) - v,$ or ${\big( {\sqrt[3]{u} - \sqrt[3]{v}}\big)}^3 + 3{\sqrt[3]{uv}}\big( {\sqrt[3]{u}}-{\sqrt[3]{v}}\big) = u - v.$

If you compare this with

 x3+px=q,

$\sqrt[3]{u} - \sqrt[3]{v}$ corresponds to $x,$ $3{\sqrt[3]{uv}}$ to $p,$ and $u-v$ to $q.$ From $3{\sqrt[3]{uv}} = p,$ you obtain $uv = \Big({\frac{p}{3}}\Big)^3,$ which, when combined with $u-v = q,$ yields quadratic equations $u^2 - qu - \Big({\frac{p}{3}}\Big)^3 = 0$ and $v^2 + qv - \Big({\frac{p}{3}}\Big)^3 = 0$ with positive solutions: $u = \frac{q}{2} + \sqrt{\Big({\frac{q}{2}}\Big)^2 + \Big({\frac{p}{3}}\Big)^3}$ $v = -\frac{q}{2} + \sqrt{\Big({\frac{q}{2}}\Big)^2 + \Big({\frac{p}{3}}\Big)^3}$

Therefore, remembering that $x = \sqrt[3]{u} - \sqrt[3]{v},$ the explicit solution of equation A) in modern notation is: $x = \sqrt[3]{ \sqrt{\Big({\frac{q}{2}}\Big)^2 + \Big({\frac{p}{3}}\Big)^3}+\frac{q}{2}} - \sqrt[3]{ \sqrt{\Big({\frac{q}{2}}\Big)^2 + \Big({\frac{p}{3}}\Big)^3}-\frac{q}{2}}.$

In the case B) x3=px+q, the path to the solution begins with cubing the binomio, the sum of two cube roots: ${\big( {\sqrt[3]{u} + \sqrt[3]{v}}\big)}^3 = 3{\sqrt[3]{uv}}\big( {\sqrt[3]{u}}+{\sqrt[3]{v}}\big) + u + v.$

Comparing this with

 x3=px+q,

you have again $uv = \Big({\frac{p}{3}}\Big)^3;$ this time, however, $u+v = q$ and $x = \sqrt[3]{u} + \sqrt[3]{v}.$

Solving the quadratic equations you get this time: $u = \frac{q}{2} + \sqrt{\Big({\frac{q}{2}}\Big)^2 - \Big({\frac{p}{3}}\Big)^3}$ $v = \frac{q}{2} - \sqrt{\Big({\frac{q}{2}}\Big)^2 - \Big({\frac{p}{3}}\Big)^3}$ provided $\Big({\frac{q}{2}}\Big)^2 \ge \Big({\frac{p}{3}}\Big)^3 .$

Therefore, the explicit solution of equation B) in modern notation is $x = \sqrt[3]{\frac{q}{2} + \sqrt{ \Big({\frac{q}{2}}\Big)^2 - \Big({\frac{p}{3}}\Big)^3}} + \sqrt[3]{\frac{q}{2} - \sqrt{\Big({\frac{q}{2}}\Big)^2 - \Big({\frac{p}{3}}\Big)^3}}.$

With the same p and q, equation C) x3 + q = px has the same solution as B), however, with opposite signs. $x = -\sqrt[3]{\frac{q}{2} + \sqrt{ \Big({\frac{q}{2}}\Big)^2 - \Big({\frac{p}{3}}\Big)^3}} - \sqrt[3]{\frac{q}{2} - \sqrt{\Big({\frac{q}{2}}\Big)^2 - \Big({\frac{p}{3}}\Big)^3}}.$ This is so because equation C) written with opposite signs, -x3+q=-px, is identical with B), x3=px+q.

Because the negative solution of C) was rejected,this type of cubic equation was not treated by most contemporary mathematicians.

Friedrich Katscher, "How Tartaglia Solved the Cubic Equation - Tartaglia's Solution in Modern Notation," Convergence (August 2011)