= Solution
A <root of unity> satisfies $|z|^3=1$, hence has <modulus> one. Writing $z=e^{i\phi}$, its <argument> satisfies $3\phi=2\pi k$ for some <integer> $k$. The three distinct <complex roots> are therefore
$$
\boxed{z=1,\qquad -\frac12+\frac{\sqrt3}{2}i,\qquad -\frac12-\frac{\sqrt3}{2}i.}
$$
They 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.
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