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: 99



Work Step by Step

Simplify the denominators to obtain: $=\dfrac{-4}{\sqrt[3]{3}} + \dfrac{1}{\sqrt[3]{8(3)}} - \dfrac{2}{\sqrt[3]{27(3)}} \\=\dfrac{-4}{\sqrt[3]{3}} + \dfrac{1}{2\sqrt[3]{3}}- \dfrac{2}{3\sqrt[3]{3}}$ Make the expressions similar by using their LCD of $6\sqrt[3]{3}$ to obtain: $=\dfrac{-4(6)}{\sqrt[3]{3}(6)}+ \dfrac{1(3)}{2\sqrt[3]{3}(3)}-\dfrac{2(2)}{3\sqrt[3]{3}(2)} \\=\dfrac{-24}{6\sqrt[3]{3}}+\dfrac{3}{6\sqrt[3]{3}}-\dfrac{4}{6\sqrt[3]{3}}$ Perform the operations to the numerators and copy the denominator to obtain: $=\dfrac{-24+3-4}{6\sqrt[3]{3}} \\=\dfrac{-25}{6\sqrt[3]{3}}$ Rationalize the denominator by multiplying $\sqrt[3]{9}$ to both the numerator and the denominator to obtain: $=\dfrac{-25(\sqrt[3]{9})}{6\sqrt[3]{3}(\sqrt[3]{9})} \\=\dfrac{-25\sqrt[3]{9}}{6(3)} \\=\color{blue}{\dfrac{-25\sqrt[3]{9}}{18}}$
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