## Intermediate Algebra: Connecting Concepts through Application

$\dfrac{81y^{8}}{x^{12}}$
$\bf{\text{Solution Outline:}}$ To simplify the given expression, $\left( \dfrac{1}{3}x^3y^{-2} \right)^{-4} ,$ 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} \left( \dfrac{1}{3}x^3y^{-2} \right)^{-4} \\\\= \left( \dfrac{1}{3}\right)^{-4}x^{3(-4)}y^{-2(-4)} \\\\= \left( \dfrac{1}{3}\right)^{-4}x^{-12}y^{8} .\end{array} Using the Power of a Quotient Rule of the laws of exponents, which is given by $\left( \dfrac{x^m}{y^n} \right)^p=\dfrac{x^{mp}}{y^{np}},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \left( \dfrac{1}{3}\right)^{-4}x^{-12}y^{8} \\\\= \dfrac{1^{-4}}{3^{-4}}x^{-12}y^{8} \\\\= \dfrac{1^{-4}\cdot x^{-12}y^{8}}{3^{-4}} .\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{1^{-4}\cdot x^{-12}y^{8}}{3^{-4}} \\\\= \dfrac{3^{4}\cdot y^{8}}{1^{4}x^{12}} \\\\= \dfrac{81y^{8}}{x^{12}} .\end{array}