21 Feb
2013
21 Feb
'13
4:52 p.m.
Duh! If I could find the affine transform to make the triangle equilateral, then I would be a single cube root away from solving the cubic. Thus, finding the affine xform is essentially equivalent to solving the _resolvent quadratic_, i.e., the penultimate step in solving the cubic. At 03:37 PM 2/21/2013, Henry Baker wrote:
Note that if the 3 triangle vertices are complex numbers that are the solution of a cubic polynomial, all of the steps up until the last one can be performed _without factoring the cubic polynomial_.