## Intermediate Algebra: Connecting Concepts through Application

$\dfrac{125a^{15}}{8b^{6}}$
$\bf{\text{Solution Outline:}}$ To simplify the given expression, $\left( \dfrac{2}{5}a^{-5}b^{2} \right)^{-3} ,$ 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{2}{5}a^{-5}b^{2} \right)^{-3} \\\\= \left( \dfrac{2}{5}\right)^{-3}a^{-5(-3)}b^{2(-3)} \\\\= \left( \dfrac{2}{5}\right)^{-3}a^{15}b^{-6} .\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{2}{5}\right)^{-3}a^{15}b^{-6} \\\\= \dfrac{2^{-3}}{5^{-3}}a^{15}b^{-6} \\\\= \dfrac{2^{-3}a^{15}b^{-6}}{5^{-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{2^{-3}a^{15}b^{-6}}{5^{-3}} \\\\= \dfrac{5^{3}a^{15}}{2^{3}b^{6}} \\\\= \dfrac{125a^{15}}{8b^{6}} .\end{array}