Zeroing the coefficient of the squared term (setting A=0) resulted in the sum of the roots being zero. There is a simple explanation for this. If we let pk=x-xk with k=0,1,2 then multiplying three monomials containing the roots together involves selecting either the unknown x or the root which is designated by 0 or 1 respectively in the table below. There are eight combinations.
There is one combination with three 0s for the x3 term, three combinations with two 0s and a 1 for the x2 term, three combinations with one 0 and two 1s for the x term and one combination with three 1s for a constant term. So when we multiply the three monomials together we get coefficients that involve the roots.
Setting the coeffient of the x2 term, A, equal to zero in the reduced polynomial also requires that the sum of the roots is also zero. If we did the same for a higher degree polynomial we would find that the sum of all the roots is also zero. Smith says that the cubic was first solved about 1515 nearly 500 years ago.