Answer
$\dfrac{75}{2c^{3}}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
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}
