According to Wikipedia and Wolfram Alpha, a cube root $n^{\frac{1}{3}}$ has three results: one real number and two complex, if $n$ is a real number; and three complex numbers if $n$ is a complex.
Most of calculators returns only the principal root, but not the other two results. Is there a way to calculate the two remaining?
On
Just multiply your root by $e^{2i\pi/3}$ and $e^{-2i\pi/3}$.
For example : solve $x^3=8$. A trivial root is 2, therefore the roots will be $$ 2\quad 2\, e^{2i\pi/3}\quad 2\, e^{-2i\pi/3}. $$ If you want to solve $x^3=-i$, a simple root is $e^{-i\pi/6}$ (this because $-i=e^{-i\pi/2}$), therefore the roots are $$ e^{-i\pi/6}\quad e^{-i\pi/6}\, e^{2i\pi/3}= e^{i\pi/2}\quad e^{-i\pi/6},\ e^{-2i\pi/3}= e^{-5i\pi/6}. $$
Yes, there is a way .e.g. First let $x^3=1$, thus $x^3-1=0$, now convert this to the form $(x-1)(x^2+x+1)=0$ solving first bracket will give you real cube root and the second one will give the imaginary cube roots.