Answer
$\frac{1}{2 b^5c^2}$
Work Step by Step
$$\begin{align*}
\left(\frac{16b^{12}c^2}{2b^{-3}c^{-4}}\right)^{-\frac{1}{3}}&=\frac{1}{\left(\frac{16b^{12}c^2}{2b^{-3}c^{-4}}\right)^{\frac{1}{3}}} \\
&=\left(\frac{1}{\frac{16b^{12}c^2}{2b^{-3}c^{-4}} }\right)^{\frac{1}{3}}\\
&=\left(\frac{2b^{-3}c^{-4}}{16b^{12}c^2}\right)^{\frac{1}{3}}\\
&=\left(\frac{2}{16b^{12}c^2b^3c^4}\right)^{\frac{1}{3}}\\
&=\left(\frac{1}{8b^{12+3}c^{2+4}}\right)^{ \frac{1}{3}}\\
&=\left(\frac{1}{8b^{15}c^6}\right)^{ \frac{1}{3}}\\
&=\left(\frac{1}{2^3 (b^5)^3 (c^2)^3}\right)^{ \frac{1}{3}}\\
&=\left(\left(\frac{1}{2 b^5c^2}\right)^3\right)^{ \frac{1}{3}}\\
&=\frac{1}{2 b^5c^2}
\end{align*}$$
