Intermediate Algebra (6th Edition)

True, the rule $\frac{1}{(a)^{-\frac{m}{n}}}=a^{\frac{m}{n}}$ (where $a^{\frac{m}{n}}$ is a nonzero real number) can be found on page 420. For example, $\frac{1}{(16)^{-\frac{3}{4}}}=16^{\frac{3}{4}}=(\sqrt[4] 16)^{3}=2^{3}=8$.