Intermediate Algebra (6th Edition)

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

Chapter 7 - Section 7.3 - Simplifying Radical Expressions - Exercise Set - Page 433: 57



Work Step by Step

$\sqrt [3](125r^{9}s^{12})=\sqrt[3] (125\times r^{9} \times s^{12})=\sqrt[3] 125\times \sqrt[3] (r^{9})\times \sqrt [3](s^{12})=5r^{3}s^{4}$ We know that $\sqrt[3] 125=5$, because $5^{3}=125$. We also know that $\sqrt[3] (r^{9})=r^{3}$, because $(r^{3})^{3}=r^{3\times3}=r^{9}$ and that $\sqrt[3] (s^{12})=s^{4}$, because $(s^{4})^{3}=s^{4\times3}=s^{12}$.
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