 Solution

A root of unity satisfies $|z{|}^3=1$, hence has modulus one. Writing $z=e^{i\varphi }$, its argument satisfies $3\varphi =2\pi k$ for some integer $k$. The three distinct complex roots are thereforeThey are equally spaced on the unit circle. Taking $k=0,1,2$ exhausts them, since adding three to $k$ repeats a root and a nonzero polynomial of degree three has at most three distinct roots.