Precalculus (6th Edition)

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

Chapter R - Review of Basic Concepts - R.7 Radical Expressions - R.7 Exercises - Page 75: 100



Work Step by Step

Simplify the denominators to obtain: $=\dfrac{5}{\sqrt[3]{2}} - \dfrac{2}{\sqrt[3]{8(2)}} + \dfrac{1}{\sqrt[3]{27(2)}} \\=\dfrac{5}{\sqrt[3]{2}} - \dfrac{2}{2\sqrt[3]{2}}+ \dfrac{1}{3\sqrt[3]{2}}$ Make the expressions similar by using their LCD of $6\sqrt[3]{2}$ to obtain: $=\dfrac{5(6)}{\sqrt[3]{2}(6)}- \dfrac{2(3)}{2\sqrt[3]{2}(3)}+\dfrac{1(2)}{3\sqrt[3]{2}(2)} \\=\dfrac{30}{6\sqrt[3]{2}}-\dfrac{6}{6\sqrt[3]{2}}+\dfrac{2}{6\sqrt[3]{2}}$ Perform the operations to the numerators and copy the denominator to obtain: $=\dfrac{30-6+2}{6\sqrt[3]{2}} \\=\dfrac{26}{6\sqrt[3]{2}}$ Rationalize the denominator by multiplying $\sqrt[3]{4}$ to both the numerator and the denominator to obtain: $=\dfrac{26(\sqrt[3]{4})}{6\sqrt[3]{2}(\sqrt[3]{4})} \\=\dfrac{26\sqrt[3]{4}}{6(\sqrt[3]{8})} \\=\dfrac{26\sqrt[3]{4}}{6(2)} \\=\dfrac{26\sqrt[3]{4}}{12}$ Cancel the common factor $2$ to obtain: $\require{cancel}=\dfrac{\cancel{26}^{13}\sqrt[3]{4}}{\cancel{12}6} \\=\color{blue}{\dfrac{13\sqrt[3]{4}}{6}}$
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