Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter R - Review of Basic Concepts - Chapter R Test Prep - Review Exercises - Page 84: 107

Answer

$\color{blue}{-\dfrac{\sqrt[3]{50p}}{5p}}$

Work Step by Step

Rationalize the denominator by multiplying $25p$ to both the numerator and the denominator of the radicand to obtain: $=-\sqrt[3]{\dfrac{2(25p)}{5p^2(25p)}} \\=-\sqrt[3]{\dfrac{50p}{125p^3}} \\=-\sqrt[3]{\dfrac{50p}{5^3p^3}}$ Bring out the cube root of the denominator to obtain: $=\color{blue}{-\dfrac{\sqrt[3]{50p}}{5p}}$
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