## Introductory Algebra for College Students (7th Edition)

Published by Pearson

# Chapter 5 - Section 5.7 - Negative Exponents and Scientific Notation - Exercise Set - Page 409: 78

#### Answer

$\dfrac{y^{8}}{x^{12}}$

#### Work Step by Step

Using the laws of exponents, the given expression, $\left( \dfrac{x^3}{y^2} \right)^{-4} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{x^{3(-4)}}{y^{2(-4)}} \\\\= \dfrac{x^{-12}}{y^{-8}} \\\\= \dfrac{y^{8}}{x^{12}} .\end{array}

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