1. Given $a-\frac{1}{a}=\frac{1}{b}-b=3$, and $a+b \neq 0$. Then, $\frac{a}{b^{3}}-\frac{b}{a^{3}}$ is ( ).
(A) $21 \sqrt{5}$
(B) $21 \sqrt{13}$
(C) $33 \sqrt{5}$
(D) $13 \sqrt{13}$
,- 1. (D).
Given $a^{2}-3 a-1=0,(-b)^{2}-3(-b)-1=0$, and $a \neq -b$. Therefore, $a$ and $(-b)$ are the roots of the equation $x^{2}-3 x-1=0$. Hence, we have
$$
\begin{array}{l}
\left\{\begin{array}{l}
a - b = 3, \\
a(-b) = -1
\end{array} \quad \text { i.e., } \left\{\begin{array}{l}
a - b = 3, \\
a b = 1
\end{array}\right.\right. \\
\therefore \frac{a}{b^{3}}-\frac{b}{a^{3}}=\frac{a^{4}-b^{4}}{(a b)^{3}} \\
=\frac{(a-b)(a+b)\left(a^{2}+b^{2}\right)}{(a b)^{3}} \\
=\frac{(a-b) \sqrt{(a-b)^{2}+4 a b}\left[(a-b)^{2}+2 a b\right]}{(a b)^{3}} \\
=\frac{3 \times \sqrt{13} \times 11}{1}=33 \sqrt{13} .
\end{array}
$$