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: 67


$2\times\sqrt[3] 4$

Work Step by Step

$\sqrt[3] 32=\sqrt[3] (8\times4)=\sqrt[3] 8\times\sqrt[3] 4=2\times\sqrt[3] 4$ We know that $\sqrt[3] 8=2$, because $2^{3}=8$.
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