#### Answer

$\frac{1}{81}$

#### Work Step by Step

We know that $a^{-\frac{m}{n}}=\frac{1}{a^{\frac{m}{n}}}$, where $a^{\frac{m}{n}}$ is a real number.
Therefore, $27^{-\frac{4}{3}}=\frac{1}{27^{\frac{4}{3}}}$.
We know that $a^{\frac{m}{n}}=\sqrt[n] a^{m}=\sqrt[n] (a)^{m}$, where all indicated roots are real numbers.
Therefore, $\frac{1}{27^{\frac{4}{3}}}=\frac{1}{\sqrt[3] 27^{4}}=\frac{1}{(\sqrt[3] 27)^{4}}=\frac{1}{3^{4}}=\frac{1}{81}$.
$\sqrt[3] 27=3$, because $3^{3}=27$