Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter R - Review of Basic Concepts - R.7 Radical Expressions - R.7 Exercises - Page 76: 115



Work Step by Step

To solve the item mentally, note that: $\sqrt[4]{8}\cdot \sqrt[4]{2} = \sqrt[4]{16}$ (because of the product rule for radicals) Since $16=2^4$. then $\sqrt[4]{16} = \sqrt[4]{2^4} = 2$
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