Intermediate Algebra (6th Edition)

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

Chapter 7 - Sections 7.1-7.5 - Integrated Review - Radicals and Rational Exponents: 28


$2a^{2}\sqrt[4] 3$

Work Step by Step

$\frac{\sqrt[4] (48a^{9}b^{3})}{\sqrt[4] (ab^{3})}=\sqrt[4] \frac {48a^{9}b^{3}}{ab^{3}}=\sqrt[4] (48a^{8})=\sqrt[4] (16\times a^{8}\times 3)=\sqrt[4] 16\times \sqrt[4] (a^{8})\times \sqrt[4] 3=2a^{2}\sqrt[4] 3$ We know that $\sqrt[4] 16=2$, because $2^{4}=16$. We know that $\sqrt[4] (a^{8})=a^{2}$, because $(a^{2})^{4}=a^{2\times4}=a^{8}$
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