## Precalculus (6th Edition) Blitzer

$\frac{1}{3}$
Recall that $a^{x/y}=\sqrt[y] {a^x}$. Also recall that the negative exponent rule states that for every nonzero number $a$ and integer $n$, $a^{-n}=\frac{1}{a^n}$. Use these rules to simplify: $27^{-1/3}=\frac{1}{27^{1/3}}=\frac{1}{\sqrt[3]{27}}=\frac{1}{3}$ (See my explanation given to exercise 54 if you've forgotten how to find a cube root.)