Ans: B
$\begin{array}{cl}
& (2n^3)^{-5} \\
= & 2^{-5} n ^{3 \times (-5)} \\
= & \dfrac{1}{2^5 n^{15}} \\
= & \dfrac{1}{32n^{15}}
\end{array}$
$\begin{array}{cl}
& (2n^3)^{-5} \\
= & 2^{-5} n ^{3 \times (-5)} \\
= & \dfrac{1}{2^5 n^{15}} \\
= & \dfrac{1}{32n^{15}}
\end{array}$
