College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.3 - Page 50: 134


False $8^{\frac{-1}{3}} = \frac{1}{2}$

Work Step by Step

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