College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter P - Prerequisites: Fundamental Concepts of Algebra - Exercise Set P.3: 113

Answer

${x^{3} y^{-2}}$ = $\frac{x^{3}}{y^{2}}$

Work Step by Step

$(\frac{x^{-\frac{5}{4}}y^{\frac{1}{3}}}{x^{-\frac{3}{4}}})^{-6}$ = $({x^{-\frac{5}{4} + \frac{3}{4}} y^{\frac{1}{3}}})^{-6}$ = $({x^{-\frac{2}{4}} y^{\frac{1}{3}}})^{-6}$ = $({x^{-\frac{2}{4}\times-6} y^{\frac{1}{3}\times-6}})$ = ${x^{3} y^{-2}}$ = $\frac{x^{3}}{y^{2}}$
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