## anonymous 4 years ago How to find the complex cube roots of -1?

1. anonymous

-1=i^2 so complex number cube root would be $\sqrt[3]{i ^{2}}$ $(i ^{2})^{1/3}$ $i ^{2/3}$ $i ^{2}-i ^{3}$ $-1-\sqrt{-1}$ but the final answer would just be until i^2/3 i think since it is asked in terms of complex numbers

2. anonymous

It may be simpler to do as follows: $x^3= -1 \rightarrow x^3+1=0 \rightarrow (x+1)(x^2-x+1)=0$ This gives x = -1 as the real root and the two complex roots can be obtained by solving the quadratic equation$x^2-x+1$