Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Section 7.1 - Radicals and Radical Functions - Exercise Set - Page 418: 98



Work Step by Step

Using the laws of exponents, the given expression, $ \dfrac{(2a^{-1}b^2)^3}{(8a^2b)^{-2}} $, simplifies to \begin{array}{l}\require{cancel} \dfrac{2^{1(3)}a^{-1(3)}b^{2(3)}}{8^{1(-2)}a^{2(-2)}b^{1(-2)}} \\\\= \dfrac{2^{3}a^{-3}b^{6}}{8^{-2}a^{-4}b^{-2}} \\\\= 2^{3}\cdot8^2a^{-3-(-4)}b^{6-(-2)} \\\\= 8\cdot64a^{-3+4}b^{6+2} \\\\= 512a^{1}b^{8} \\\\= 512a^{}b^{8} .\end{array}
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