## College Algebra (6th Edition)

False $8^{\frac{-1}{3}} = \frac{1}{2}$
By the definition of $a^{\frac{-1}{n}} = \frac{1}{a^{\frac{1}{n}}} = \frac{1}{\sqrt[3]a}$ where $a \ne 0$ $8^{\frac{-1}{3}} = \frac{1}{8^{\frac{1}{3}}}$ $= \frac{1}{\sqrt[3]8} = \frac{1}{2}$