Intermediate Algebra (6th Edition)

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

Chapter 7 - Review - Page 468: 9



Work Step by Step

$\sqrt[3] (-a^{6}b^{9})=\sqrt[3] (-1\times a^{6}\times b^{9})=\sqrt[3] (-1) \times \sqrt[3] (a^{6})\times \sqrt[3] (b^{9})=-a^{2}b^{3}$ $\sqrt[3] (-1)=-1$, because $(-1)^{3}=-1\times-1\times-1=-1$. $\sqrt[3] (a^{6})=a^{2}$, because $(a^{2})^{3}=a^{2\times3}=a^{6}$. $\sqrt[3] (b^{9})=b^{3}$, because $(b^{3})^{3}=b^{3\times3}=b^{9}$.
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