Solving for the roots of a cubic polynomial

[Albert Chan]

(07-01-2024 04:13 AM)Namir Wrote:  

(06-30-2024 11:49 PM)Albert Chan Wrote:  Try its depressed cubic, with x = y - 1/3

y^3 - y/3 + 2/27 = 0

Sure enough, I get d=0, X1 = X2 = X3 = 0

I think the HP author of the program assumed that parameter c should not be 0, because in this case the cubic equation s easily reduced to a quadratic or linear equation (as is your example).

Depressed cubic does not look easily reduced to quadratic.
Fix should not be hard. Here is the trig solution.