Cardano’s derivation of the cubic formula

To solve the cubic polynomial equation x3+ax2+bx+c=0 for x, the first step is to apply the Tchirnhaus transformation x=y-a3. This reduces the equation to y3+py+q=0, where

p = b-a23
q = c-ab3+2a327

The next step is to substitute y=u-v, to obtain

(u-v)3+p(u-v)+q=0 (1)

or, with the terms collected,

(q-(v3-u3))+(u-v)(p-3uv)=0 (2)

From equation (2), we see that if u and v are chosen so that q=v3-u3 and p=3uv, then y=u-v will satisfy equation (1), and the cubic equationMathworldPlanetmath will be solved!

There remains the matter of solving q=v3-u3 and p=3uv for u and v. From the second equation, we get v=p/(3u), and substituting this v into the first equation yields


which is a quadratic equation in u3. Solving for u3 using the quadratic formula, we get

u3 = -27q+108p3+729q254=-9q+12p3+81q218
v3 = 27q+108p3+729q254=9q+12p3+81q218

Using these values for u and v, you can back–substitute y=u-v, p=b-a2/3, q=c-ab/3+2a3/27, and x=y-a/3 to get the expression for the first root r1 in the cubic formula. The second and third roots r2 and r3 are obtained by performing synthetic divisionMathworldPlanetmath using r1, and using the quadratic formula on the remaining quadratic factor.

Title Cardano’s derivation of the cubic formula
Canonical name CardanosDerivationOfTheCubicFormula
Date of creation 2013-03-22 12:10:28
Last modified on 2013-03-22 12:10:28
Owner djao (24)
Last modified by djao (24)
Numerical id 12
Author djao (24)
Entry type Proof
Classification msc 12D10
Related topic FerrariCardanoDerivationOfTheQuarticFormula