Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 4 - Section 4.1 - Integer Exponents and Scientific Notation - 4.1 Exercises - Page 278: 144



Work Step by Step

Using the laws of exponents, the given expression, $ \left( \dfrac{-4a^3b^2}{12a^5b^{-4}} \right)^{-3} ,$ is equivalent to \begin{array}{l}\require{cancel} \dfrac{(-4)^{-3}a^{3(-3)}b^{2(-3)}}{12^{-3}a^{5(-3)}b^{-4(-3)}} \\\\= \dfrac{(-4)^{-3}a^{-9}b^{-6}}{12^{-3}a^{-15}b^{12}} \\\\= \dfrac{(-4)^{-3}a^{-9-(-15)}b^{-6-12}}{(-4)^{-3}(-3)^{-3}} \\\\= \dfrac{\cancel{(-4)^{-3}}a^{-9+15}b^{-6-12}}{\cancel{(-4)^{-3}}(-3)^{-3}} \\\\= \dfrac{a^{6}b^{-18}}{(-3)^{-3}} \\\\= \dfrac{(-3)^{3}a^{6}}{b^{18}} \\\\= \dfrac{-27a^{6}}{b^{18}} \\\\= -\dfrac{27a^{6}}{b^{18}} .\end{array}
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