Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 10 - Exponents and Radicals - Review Exercises: Chapter 10: 28

Answer

$-6x^{5}y^{4}\sqrt[3]{2x^2}$

Work Step by Step

Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ the given expression is equivalent to \begin{array}{l}\require{cancel} \sqrt[3]{-24x^{10}y^8}\sqrt[3]{18x^7y^4} \\\\= \sqrt[3]{(-8\cdot3)(2\cdot9)x^{10+7}y^{8+4}} \\\\= \sqrt[3]{(-8\cdot27\cdot2)x^{17}y^{12}} .\end{array} Extracting the factors that are perfect powers of the index, the given expression simplifies to \begin{array}{l}\require{cancel} \sqrt[3]{(-8\cdot27\cdot2)x^{17}y^{12}} \\\\= \sqrt[3]{(-8\cdot27)x^{15}y^{12}\cdot2x^2} \\\\= \sqrt[3]{[(-2\cdot3)x^{5}y^{4}]^3\cdot2x^2} \\\\= \sqrt[3]{[(-2\cdot3)x^{5}y^{4}]^3\cdot2x^2} \\\\= (-2\cdot3)x^{5}y^{4}\sqrt[3]{2x^2} \\\\= -6x^{5}y^{4}\sqrt[3]{2x^2} .\end{array}
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