3. If $a$ is a complex number, and $\left(a^{2}-a\right)^{2}=1$, then $a^{3}-2 a^{2}$ can take ( ) different values.
(A)2
(B) 3
(C) 4
(D) 6
3. B.

From the given, we have $a^{4}-2 a^{3}+a^{2}=1$, so $a^{3}-2 a^{2}=\frac{1-a^{2}}{a}($ obviously $a \neq 0)$. When $a^{2}-a=1$, $1-a^{2}=-a$, then $a^{3}-2 a^{2}=-1$; when $a^{2}-a=-1$, $a^{2}=a-1, a=\frac{1 \pm \sqrt{3} \mathrm{i}}{2}$, then $a^{3}-2 a^{2}=\frac{1-(a-1)}{a}=-1+\frac{2}{a}$.
Thus, $a^{3}-2 a^{2}$ can take two different imaginary values, i.e., $a^{3}-2 a^{2}= \pm \sqrt{3}$ i.
In summary, $a^{3}-2 a^{2}$ can take a total of 3 complex values.