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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.

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