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The solution of the cubic

What is now called ``Cardan's method" to solve the cubic, but that should be more correctly called ``del Ferro's method" goes thus: consider the general equation of third degree tex2html_wrap_inline172. A substitution tex2html_wrap_inline174 gathers tex2html_wrap_inline176. Let tex2html_wrap_inline178. This implies that solving the above cubic is equivalent to solving the cubic tex2html_wrap_inline180 where tex2html_wrap_inline182 and tex2html_wrap_inline184. Now substitute t=u+v. This means that tex2html_wrap_inline188. The latter is satisfied when tex2html_wrap_inline190 and 3uv+p=0. Hence tex2html_wrap_inline194 and tex2html_wrap_inline196, so tex2html_wrap_inline198 and tex2html_wrap_inline200 are the roots of the quadratic tex2html_wrap_inline202. Hence one can find u and v and therefore t=u+v in terms of p and q. The quartic had been solved with a method not much different from the one above.

Leo Liberti
Thu Feb 26 17:04:11 CET 1998