$12 \cdot 6$ Let non-zero complex numbers $x, y$ satisfy $x^{2}+x y+y^{2}=0$, then the value of the algebraic expression $\left(\frac{x}{x+y}\right)^{1990}+\left(\frac{y}{x+y}\right)^{1990}$ is
(A) $2^{-1989}$.
(B) -1 .
(C) 1 .
(D) None of the above answers is correct.
(China High School Mathematics League, 1990)
[Solution] Let $y=\omega x, \omega \neq 1$, substituting into the given condition yields $1+\omega+\omega^{2}=0$. Therefore, $(1-\omega)\left(1+\omega+\omega^{2}\right)=0$, which implies $\omega^{3}=1$.
$$
\begin{aligned}
\therefore \quad \text { the original expression } & =\frac{1}{(1+\omega)^{1990}}+\frac{\omega^{1990}}{(1+\omega)^{1990}}=\frac{1+\omega^{1990}}{\left(-\omega^{2}\right)^{1990}} \\
& =\frac{1+\omega^{3 \times 663+1}}{\omega^{2 \times(3 \times 663+1)}}=\frac{1+\omega}{\omega^{2}}=-1 .
\end{aligned}
$$

Therefore, the answer is $(B)$.