Finite Math and Applied Calculus (6th Edition)

Published by Brooks Cole
ISBN 10: 1133607705
ISBN 13: 978-1-13360-770-0

Chapter 0 - Section 0.2 - Exponents and Radicals - Exercises - Page 17: 100

Answer

$\frac{1}{\sqrt[3] 2}$

Work Step by Step

$2^{\frac{1}{3}}2^{-1}2^{\frac{2}{3}}2^{-\frac{1}{3}}=2^{\frac{1}{3}-1}2^{\frac{2}{3}-\frac{1}{3}}=2^{-\frac{2}{3}}2^{\frac{1}{3}}=2^{-\frac{2}{3}+\frac{1}{3}}=2^{-\frac{1}{3}}=\frac{1}{2^{\frac{1}{3}}}=\frac{1}{\sqrt[3] 2}$
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