## College Algebra (11th Edition)

$\dfrac{256}{81}$
$\bf{\text{Solution Outline:}}$ Use the laws of exponents to simplify the given expression, $\left( \dfrac{27}{64} \right)^{-4/3} .$ $\bf{\text{Solution Details:}}$ Using the extended Power Rule of the laws of exponents which states that $\left( \dfrac{x^m}{z^p} \right)^q=\dfrac{x^{mq}}{z^{pq}},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \dfrac{27^{-\frac{4}{3}}}{64^{-\frac{4}{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{64^{\frac{4}{3}}}{27^{\frac{4}{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]{64})^4}{(\sqrt[3]{27})^4} \\\\= \dfrac{(\sqrt[3]{4^3})^4}{(\sqrt[3]{3^3})^4} \\\\= \dfrac{(4)^4}{(3)^4} \\\\= \dfrac{256}{81} .\end{array}