Intermediate Algebra (6th Edition)

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

Chapter 9 - Cumulative Review - Page 601: 32

Answer

$\dfrac{3\sqrt[3]{m^2n}}{m^2n^3}$

Work Step by Step

Rationalizing the denominator of $ \sqrt[3]{\dfrac{27}{m^4n^8}} $ results to \begin{array}{l} \sqrt[3]{\dfrac{27}{m^3n^6}\cdot\dfrac{1}{mn^2}} \\\\= \dfrac{3}{mn^2}\sqrt[3]{\dfrac{1}{mn^2}} \\\\= \dfrac{3}{mn^2}\sqrt[3]{\dfrac{1}{mn^2}\cdot\dfrac{m^2n}{m^2n}} \\\\= \dfrac{3}{mn^2}\sqrt[3]{\dfrac{m^2n}{m^3n^3}} \\\\= \dfrac{3\sqrt[3]{m^2n}}{mn^2\cdot mn} \\\\= \dfrac{3\sqrt[3]{m^2n}}{m^2n^3} .\end{array}
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