Why can't a cubic equation have 3 equal roots?

It absolutely can. Consider, for instance, [math](x-a)^3=0[/math] as an equation in [math]x[/math]. All of its roots are equal to [math]a[/math], and it is a cubic equation in [math]x[/math]. read more

As the others have pointed out, a cubic equation can indeed have three equal roots. What it cannot have, however, is three equal complex-valued roots. This is because polynomials having real-valued coefficients can only have complex-valued roots in conjugate pairs. read more

Since every real number can be thought of as a complex number, one way to answer this is to say the answer is 3, after"counting multiplicities". On the other hand, if you are after non-real complex roots, and if the coefficients of your cubic equation are real numbers, the answer will be 0 or 2. read more