12.5 If $\omega$ is a complex root of the equation $x^{3}=1$, then the product $\left(1-\omega+\omega^{2}\right)$ $\left(1+\omega-\omega^{2}\right)$ equals
(A) 4 .
(B) $\omega$.
(C) 2 .
(D) $\omega^{2}$.
(E) 1 .
(22nd American High School Mathematics Examination, 1971)
[Solution] Factorizing the given equation $x^{3}-1=0$, we get
$$
(x-1)\left(x^{2}+x+1\right)=0 .
$$

Its imaginary root $\omega$ satisfies $\quad \omega^{2}+\omega+1=0, \quad$ and $\omega^{3}=1$.
Therefore, $\quad 1+\omega^{2}=-\omega, \quad 1+\omega=-\omega^{2}$.
$$
\therefore \quad\left(1-\omega+\omega^{2}\right)\left(1+\omega-\omega^{2}\right)=(-2 \omega)\left(-2 \omega^{2}\right)=4 \omega^{3}=4 \text {. }
$$

Thus, the answer is $(A)$.