# Tartaglia-Cardano Derivation of the Cubic Formula

A solution is sought for an equation in the form

x^3 + ax = b, a \gt 0, b \gt 0.

By direct verification,

(p - q)^3 + 3pq(p - q) = p^3 - q^3.

Therefore, letting x = p - q  we may also identify a = 3pq  and b = p^3 - q^3.

27(q^3)^2 + 27bq^3 - a^3=0

solving which gives

q^3 = \frac{-b}{2} \pm \sqrt{\frac{b^2}{4} + \frac{a^3}{27}}

so that

q = \sqrt[3]{\frac{-b}{2} + \sqrt{\frac{b^2}{4} + \frac{a^3}{27}}}.

where the possibility of a negative q has been discarded as an unacceptable oddity. Quite similarly we can find that

p = \sqrt[3]{\frac{b}{2} + \sqrt{\frac{b^2}{4} + \frac{a^3}{27}}}.

Thus

x = p - q = \sqrt[3]{\frac{b}{2} + \sqrt{\frac{b^2}{4} + \frac{a^3}{27}}} - \sqrt[3]{\frac{-b}{2} + \sqrt{\frac{b^2}{4} + \frac{a^3}{27}}}.

Note that in the equation x^3 + ax = b  both coefficients were assumed positive. Should a   happen to be negative, the mathematicians of the 16th century would solve x^3 = ax + b  instead. The reason for this becomes clear from even a superficial inspection of the original Ars Magna.

### References

1. F. J. Swetz, From Five Fingers to Infinity'', Open Court, 1996 (Third printing), p. 368