$1 \cdot 33$ If $|x-8 y|+(4 y-1)^{2}=0$, then the value of $(x+2 y)^{3}$ is
(A) $6 \frac{1}{4}$.
(B) $11 \frac{25}{64}$.
(C) 27 .
(D) $15 \frac{5}{8}$.
(China Beijing Junior High School Mathematics Competition, 1982)
[Solution] From $|x-8 y|+(4 y-1)^{2}=0$, we have $\left\{\begin{array}{l}x-8 y=0, \\ 4 y-1=0,\end{array}\right.$ solving this gives $\left\{\begin{array}{l}x=2, \\ y=\frac{1}{4} .\end{array}\right.$ $\therefore(x+2 y)^{3}=\left(2+\frac{1}{2}\right)^{3}=\left(\frac{5}{2}\right)^{3}=15 \frac{5}{8}$. Therefore, the answer is $(D)$.