Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Review Exercises - Page 143: 61

Answer

$13 \sqrt[3]{2}$

Work Step by Step

$\sqrt[3]{16}= \sqrt[3]{2^{4}}= \sqrt[3]{2^{3}\cdot 2}=2 \sqrt[3]{2}$ Which we use in the problem as follows: $4\sqrt[3]{16}+5 \sqrt[3]{2} = 4(2 \sqrt[3]{2})+5 \sqrt[3]{2}$ $=8 \sqrt[3]{2}+5 \sqrt[3]{2}$ = $13 \sqrt[3]{2}$
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