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: 92

Answer

$\dfrac{13\sqrt[3]{4}}{6}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ Simplify each term in the given expression, $ \dfrac{5}{\sqrt[3]{2}}-\dfrac{2}{\sqrt[3]{16}}+\dfrac{1}{\sqrt[3]{54}} .$ Then make the terms similar (same denominator) to combine the numerators. $\bf{\text{Solution Details:}}$ Extracting the perfect cube factors of the radicals results to \begin{array}{l}\require{cancel} \dfrac{5}{\sqrt[3]{2}}-\dfrac{2}{\sqrt[3]{8\cdot2}}+\dfrac{1}{\sqrt[3]{27\cdot2}} \\\\= \dfrac{5}{\sqrt[3]{2}}-\dfrac{2}{\sqrt[3]{(2)^3\cdot2}}+\dfrac{1}{\sqrt[3]{(3)^3\cdot2}} \\\\= \dfrac{5}{\sqrt[3]{2}}-\dfrac{2}{2\sqrt[3]{2}}+\dfrac{1}{3\sqrt[3]{2}} \\\\= \dfrac{5}{\sqrt[3]{2}}-\dfrac{\cancel2}{\cancel2\sqrt[3]{2}}+\dfrac{1}{3\sqrt[3]{2}} \\\\= \dfrac{5}{\sqrt[3]{2}}-\dfrac{1}{\sqrt[3]{2}}+\dfrac{1}{3\sqrt[3]{2}} .\end{array} Rationalizing the denominators results to \begin{array}{l}\require{cancel} \dfrac{5}{\sqrt[3]{2}}\cdot\dfrac{\sqrt[3]{2^2}}{\sqrt[3]{2^2}}-\dfrac{1}{\sqrt[3]{2}}\cdot\dfrac{\sqrt[3]{2^2}}{\sqrt[3]{2^2}}+\dfrac{1}{3\sqrt[3]{2}}\cdot\dfrac{\sqrt[3]{2^2}}{\sqrt[3]{2^2}} \\\\= \dfrac{5\sqrt[3]{2^2}}{\sqrt[3]{2^3}}-\dfrac{\sqrt[3]{2^2}}{\sqrt[3]{2^3}}+\dfrac{\sqrt[3]{2^2}}{3\sqrt[3]{2^3}} \\\\= \dfrac{5\sqrt[3]{4}}{2}-\dfrac{\sqrt[3]{4}}{2}+\dfrac{\sqrt[3]{4}}{3(2)} \\\\= \dfrac{5\sqrt[3]{4}}{2}-\dfrac{\sqrt[3]{4}}{2}+\dfrac{\sqrt[3]{4}}{6} .\end{array} To simplify the expression above, find the $LCD$ of the denominators. The $LCD$ is $6$ since it is the lowest number that can be exactly divided by the denominators $ 2 \text{ and } 6 .$ Multiplying each term by an expression equal to $1$ so that its denominator becomes the $LCD$ results to \begin{array}{l}\require{cancel} \dfrac{5\sqrt[3]{4}}{2}\cdot\dfrac{3}{3}-\dfrac{\sqrt[3]{4}}{2}\cdot\dfrac{3}{3}+\dfrac{\sqrt[3]{4}}{6} \\\\= \dfrac{15\sqrt[3]{4}}{6}-\dfrac{3\sqrt[3]{4}}{6}+\dfrac{\sqrt[3]{4}}{6} .\end{array} To combine similar terms, add/subtract the numerators and copy the similar denominator. Hence, the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{15\sqrt[3]{4}-3\sqrt[3]{4}+\sqrt[3]{4}}{6} \\\\= \dfrac{13\sqrt[3]{4}}{6} .\end{array}
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