## Intermediate Algebra (12th Edition)

$\dfrac{25}{16}$
$\bf{\text{Solution Outline:}}$ Use the laws of exponents and the definition of rational exponents to simplify the given expression, $\left( \dfrac{64}{125} \right)^{-2/3} .$ $\bf{\text{Solution Details:}}$ Using the extended Power Rule of the laws of exponents which states that $\left( \dfrac{x^my^n}{z^p} \right)^q=\dfrac{x^{mq}y^{nq}}{z^{pq}},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{64^{-2/3}}{125^{-2/3}} .\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{125^{2/3}}{64^{2/3}} .\end{array} Using the definition of rational exponents which is given by $a^{\frac{m}{n}}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m,$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{(\sqrt[3]{125})^{2}}{(\sqrt[3]{64})^{2}} \\\\= \dfrac{(5)^{2}}{(4)^{2}} \\\\= \dfrac{25}{16} .\end{array}