Answer
$\frac{1}{16}$
Work Step by Step
We know that $(a)^{-\frac{m}{n}}=\frac{1}{(a)^{\frac{m}{n}}}$ (as long as $(a)^{\frac{m}{n}}$ is a real number).
Therefore, $8^{-\frac{4}{3}}=\frac{1}{8^{\frac{4}{3}}}$
We also know that $(a)^{\frac{m}{n}}=(\sqrt[n] a)^{m}$ (where m and n are positive integers greater than 1 with $\frac{m}{n}$ in simplest form and $\sqrt[n] a$ is a real number).
$\frac{1}{8^{\frac{4}{3}}}=\frac{1}{(\sqrt[3] 8)^{4}}=\frac{1}{2^{4}}=\frac{1}{2\times2\times2\times2}=\frac{1}{16}$
We know that $\sqrt[3] 8=2$, because $2^{3}=8$
