College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter R - Test: 29

Answer

$\dfrac{y}{x}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To simplify the given expression, $ \left( \dfrac{x^{-2}y^{-1/3}}{x^{-5/3}y^{-2/3}} \right)^3 ,$ use the laws of exponents. $\bf{\text{Solution Details:}}$ Using the extended Power Rule of the laws of exponents which is given by $\left( x^my^n \right)^p=x^{mp}y^{np},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{x^{-2(3)}y^{-\frac{1}{3}\cdot3}}{x^{-\frac{5}{3}\cdot3}y^{-\frac{2}{3}\cdot3}} \\\\= \dfrac{x^{-6}y^{-1}}{x^{-5}y^{-2}} .\end{array} Using the Quotient Rule of the laws of exponents which states that $\dfrac{x^m}{x^n}=x^{m-n},$ the expression above simplifies to \begin{array}{l}\require{cancel} x^{-6-(-5)}y^{-1-(-2)} \\\\= x^{-6+5}y^{-1+2} \\\\= x^{-1}y^{1} .\end{array} Using the Negative Exponent Rule of the laws of exponents which states that $x^{-m}=\dfrac{1}{x^m}$ or $\dfrac{1}{x^{-m}}=x^m,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{y^{1}}{x^{1}} \\\\= \dfrac{y}{x} .\end{array}
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