Has anyone seen the following treatment of the real cubic? A cubic equation with 3 real zeros can be reduced to either (your choice): 4*c^3-3*c=c3 or 4*s^3-3*s=-s3 where c,c3 reminds us of cos(a),cos(3*a), and s,s3 reminds us of sin(b),sin(3*b), respectively. Now Newton's method _can_ be used to compute one of the zeros, but an unfortunate choice of starting point will attempt to converge on a double (or almost double) root, thereby seriously slowing the convergence. For example, Newton on the cosine version gives c' = (c3+8*c^3)/[3*(2*c-1)*(2*c+1)] which leads to very unfortunate behavior for c near +- 1/2. But if we recognize that we are attempting to solve the 1-dimensional projection of a 2-dimensional problem, then we can avoid this convergence problem. Specifically, if we are given 4*c^3-3*c=c3, then we first compute s3=sqrt(1-c3^2), and then instead solve the problem z^3 = z3 = = c3+i*s3 = (c+i*s)^3 = c3+i*s3 = c^3 - 3*c*s^2 - i*(s^3 - 3*s*c^2) = c^3 - 3*c*(1-c^2) - i*(s^3 - 3*s*(1-s^2)) = c^3 - 3*c +3*c^3 - i*(s^3 - 3*s + 3*s^3) = 4*c^3 - 3*c - i*(4*s^3 - 3*s) = c3 - i*(-s3) = c3 + i*s3 But Newton's method on the complex problem z^3=z3 is much more robust, so long as we stay away from z=0: z' = (z3+2*z^3)/(3*z^2) (Throw away s=imagpart(z) & keep c=realpart(z) for the answer. But before doing so, we can trivially produce the other two roots with complex multiplications: w*z and w^2*z, where w is cube root of unity: w^3=1.) But we know that |z|^2 must be 1, so we can easily avoid the region near z=0. On modern hardware with multiple floating point units, complex arithmetic may have no more latency than real arithmetic, so this method may be no slower, and (due to faster convergence) may actually be faster.