Answer
$\frac{1}{64a^{6}}$
Work Step by Step
We know that $a^{−n}=\frac{1}{a^{n}}$ and $\frac{1}{a^{-n}}=a^{n}$ (as long as a is a nonzero real number and n is an integer).
Therefore, $(4a^{2})^{-3}=4^{-3}(a^{2})^{-3}=\frac{1}{4^{3}}\times a^{2\times-3}=\frac{1}{64}\times a^{-6}=\frac{1}{64}\times\frac{1}{a^{6}}=\frac{1}{64a^{6}}$.