Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Appendix A - Review - A.10 nth Roots; Rational Exponents - A.10 Assess Your Understanding - Page A88: 59

Answer

$ \dfrac{5\sqrt[3] 4}{2}$

Work Step by Step

Rationalize the denominator by multiplying $\sqrt[3]{4}$ to both the numerator and the denominator: $\dfrac{5}{\sqrt[3] 2}=\dfrac{5}{\sqrt[3] 2}\cdot \dfrac{\sqrt[3] 4}{\sqrt[3] 4}$ Use the rule $\sqrt[n]{a} \cdot \sqrt[n]{b}=\sqrt[n]{a\cdot b}$ then simplify: $= \dfrac{5\sqrt[3] 4}{\sqrt[3] {2\cdot 4}}$ $= \dfrac{5\sqrt[3] 4}{\sqrt[3] {8}}$ $= \dfrac{5\sqrt[3] 4}{\sqrt[3] {2^3}}$ $= \dfrac{5\sqrt[3] 4}{2}$ Hence, the correct answer is $ \frac{5\sqrt[3] 4}{2}$.
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