## College Algebra (11th Edition)

$\dfrac{9}{16}$
$\bf{\text{Solution Outline:}}$ To simplify the given expression, $\left( -\dfrac{64}{27} \right)^{-2/3} ,$ use the laws of exponents. $\bf{\text{Solution Details:}}$ Using the extended Power Rule of the laws of exponents which states that $\left( \dfrac{y^n}{z^p} \right)^q=\dfrac{y^{nq}}{z^{pq}},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \left( \dfrac{-64}{27} \right)^{-2/3} \\\\= \dfrac{(-64)^{-2/3}}{(27)^{-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{(27)^{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]{27})^{2}}{(\sqrt[3]{-64})^{2}} \\\\= \dfrac{(\sqrt[3]{3^3})^2}{\sqrt[3]{(-4)^3})^2} \\\\= \dfrac{(3)^{2}}{(-4)^{2}} \\\\= \dfrac{9}{16} .\end{array}