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 . A substitution
gathers . Let . This implies that solving the above cubic is equivalent to
solving the cubic where and . Now substitute *t*=*u*+*v*. This means that
. The latter is satisfied when and
3*uv*+*p*=0. Hence and , so and
are the roots of the quadratic . 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.

Thu Feb 26 17:04:11 CET 1998