$1 \cdot 39$ If $\left(a+\frac{1}{3 a}\right)^{2}=3$, then $27 a^{3}+\frac{1}{a^{3}}$ equals
(A) 0.
(B) $54 \sqrt{3}$.
(C) $\pm 27 \sqrt{3}$.
(D) $\pm 54 \sqrt{3}$.
(1st "Five Sheep Cup" Junior High School Mathematics Competition, 1989)
[Solution]
$$
\begin{aligned}
\text { Original expression } & =27\left(a^{3}+\frac{1}{27 a^{3}}\right)=27\left[\left(a+\frac{1}{3 a}\right)^{3}-\left(a+\frac{1}{3 a}\right)\right] \\
& =27\left(a+\frac{1}{3 a}\right)\left[\left(a+\frac{1}{3 a}\right)^{2}-1\right] \\
& =27 \cdot( \pm \sqrt{3}) \cdot[3-1] \\
& = \pm 54 \sqrt{3} .
\end{aligned}
$$

Therefore, the answer is $(D)$.