College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter R - Section R.7 - Radical Expressions - R.7 Exercises: 60

Answer

$\dfrac{\sqrt[3]{36p^2}}{4p^2} $

Work Step by Step

$\bf{\text{Solution Outline:}}$ To simplify the given expression, $ \sqrt[3]{\dfrac{9}{16p^4}} ,$ make the denominator a perfect power of the index so that the final result will already be in rationalized form. Then find a factor of the radicand that is a perfect power of the index. Finally, extract the root of that factor. $\bf{\text{Solution Details:}}$ Multiplying the radicand that will make the denominator a perfect power of the index results to \begin{array}{l}\require{cancel} \sqrt[3]{\dfrac{9}{16p^4}\cdot\dfrac{4p^2}{4p^2}} \\\\= \sqrt[3]{\dfrac{36p^2}{64p^6}} .\end{array} Factoring the radicand into an expression that is a perfect power of the index and then extracting its root result to \begin{array}{l}\require{cancel} \sqrt[3]{\dfrac{1}{64p^6}\cdot36p^2} \\\\= \sqrt[3]{\left(\dfrac{1}{4p^2}\right)^3\cdot36p^2} \\\\= \dfrac{1}{4p^2}\sqrt[3]{36p^2} \\\\= \dfrac{\sqrt[3]{36p^2}}{4p^2} .\end{array} Note that all variables are assumed to have positive values.
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