Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Review - Page 469: 68


$2ab^{2}\sqrt[3] (3a^{2}b)$

Work Step by Step

$\sqrt[3] (24a^{5}b^{7})=\sqrt[3] (8\times a^{3}\times b^{6}\times3a^{2}b)=\sqrt[3] 8\times\sqrt[3] (a^{3})\times\sqrt [3] (b^{6})\times\sqrt[3] (3a^{2}b)=2ab^{2}\sqrt[3] (3a^{2}b)$ $\sqrt[3] 8=2$, because $2^{3}=8$ $\sqrt[3] (a^{3})=a$, because $(a)^{3}=a^{3}$ $\sqrt[3] (b^{6})=b^{2}$, because $(b^{3})^{2}=b^{3\times2}=b^{6}$
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