## College Algebra (6th Edition)

$\frac{1}{81}$
We know that $b^{-n}=\frac{1}{b^{n}}$ (where $b$ is a nonzero real number and $n$ is a natural number). Therefore, $27^{-\frac{4}{3}}=\frac{1}{27^{\frac{4}{3}}}$ Based on the definition of $a^{\frac{m}{n}}$, we know that $a^{\frac{m}{n}}=(\sqrt[n] a)^{m}=\sqrt[n] a^{m}$ (where $\sqrt[n] a$ is a real number). Therefore, $\frac{1}{27^{\frac{4}{3}}}=\frac{1}{\sqrt[3] 27^{4}}=\frac{1}{(\sqrt[3] 27)^{4}}=\frac{1}{(3)^{4}}=\frac{1}{81}$ We know that $\sqrt[3] 27=3$, because $3^{3}=27$.