## College Algebra (11th Edition)

$\dfrac{1}{p^{2}(m+n)}$
$\bf{\text{Solution Outline:}}$ Use the laws of exponents to simplify the given expression, $\dfrac{[p^2(m+n)^3]^{-2}}{p^{-2}(m+n)^{-5}} .$ $\bf{\text{Solution Details:}}$ 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} \dfrac{p^{2(-2)}(m+n)^{3(-2)}}{p^{-2}(m+n)^{-5}} \\\\= \dfrac{p^{-4}(m+n)^{-6}}{p^{-2}(m+n)^{-5}} .\end{array} Using the Quotient Rule of the laws of exponents which states that $\dfrac{x^m}{x^n}=x^{m-n},$ the expression above simplifies to \begin{array}{l}\require{cancel} p^{-4-(-2)}(m+n)^{-6-(-5)} \\\\= p^{-4+2}(m+n)^{-6+5} \\\\= p^{-2}(m+n)^{-1} .\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{1}{p^{2}(m+n)^{1}} \\\\= \dfrac{1}{p^{2}(m+n)} .\end{array}