#### Answer

$\dfrac{9}{25}$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
Use the laws of exponents and the definition of rational exponents to simplify the given expression, $
\left( \dfrac{125}{27} \right)^{-2/3}
.$
$\bf{\text{Solution Details:}}$
Using the extended Power Rule of the laws of exponents which states that $\left( \dfrac{x^my^n}{z^p} \right)^q=\dfrac{x^{mq}y^{nq}}{z^{pq}},$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{125^{-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}}{125^{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]{125})^{2}}
\\\\=
\dfrac{(3)^{2}}{(5)^{2}}
\\\\=
\dfrac{9}{25}
.\end{array}