Intermediate Algebra: Connecting Concepts through Application

$\dfrac{75}{2c^{3}}$
$\bf{\text{Solution Outline:}}$ To simplify the given expression, $\left( \dfrac{1}{5}a^{-2}b^3c \right)^{-2}\left( \dfrac{2}{3}a^{4}b^{-6}c \right)^{-1} ,$ use the laws of exponents. $\bf{\text{Solution Details:}}$ 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}{5}a^{-2}b^3c \right)^{-2}\left( \dfrac{2}{3}a^{4}b^{-6}c \right)^{-1} \\\\= \left( \dfrac{a^{-2}b^3c}{5} \right)^{-2}\left( \dfrac{2a^{4}b^{-6}c}{3} \right)^{-1} \\\\= \left( \dfrac{a^{-2(-2)}b^{3(-2)}c^{-2}}{5^{-2}} \right)\left( \dfrac{2^{-1}a^{4(-1)}b^{-6(-1)}c^{-1}}{3^{-1}} \right) \\\\= \left( \dfrac{a^{4}b^{-6}c^{-2}}{5^{-2}} \right)\left( \dfrac{2^{-1}a^{-4}b^{6}c^{-1}}{3^{-1}} \right) .\end{array} Using the Product Rule of the laws of exponents, which is given by $x^m\cdot x^n=x^{m+n},$ the expression above is equivalent to \begin{array}{l}\require{cancel} \left( \dfrac{a^{4}b^{-6}c^{-2}}{5^{-2}} \right)\left( \dfrac{2^{-1}a^{-4}b^{6}c^{-1}}{3^{-1}} \right) \\\\= \dfrac{2^{-1}a^{4+(-4)}b^{-6+6}c^{-2+(-1)}}{5^{-2}\left(3^{-1}\right)} \\\\= \dfrac{2^{-1}a^{0}b^{0}c^{-3}}{5^{-2}\left(3^{-1}\right)} \\\\= \dfrac{2^{-1}(1)(1)c^{-3}}{5^{-2}\left(3^{-1}\right)} \\\\= \dfrac{2^{-1}c^{-3}}{5^{-2}\left(3^{-1}\right)} .\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^{-1}c^{-3}}{5^{-2}\left(3^{-1}\right)} \\\\= \dfrac{5^{2}\left(3^{1}\right)}{2^{1}c^{3}} \\\\= \dfrac{25(3)}{2c^{3}} \\\\= \dfrac{75}{2c^{3}} .\end{array}