Answer
(a) $\frac{1}{4y^{2}}$
(b) $\frac{1}{u^{4/3}v^{2}}$
Work Step by Step
(a) $(8y^{3})^{-2/3}=8^{-2/3}y^{3*-2/3}=\frac{1}{4y^{2}}$
(b) $(u^{4}v^{6})^{-1/3}=u^{4*-1/3}v^{6*-1/3}=\frac{1}{u^{4/3}v^{2}}$
College Algebra 7th Edition
by Stewart, James; Redlin, Lothar; Watson, Saleem
Published by
Brooks Cole
ISBN 10:
1305115546
ISBN 13:
978-1-30511-554-5
Chapter P, Prerequisites - Section P.4 - Rational Exponents and Radicals - P.4 Exercises - Page 30: 65
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