Chapter 5 - Exponents and Polynomials - 5.6 - Integral Exponents and Scientific Notation - Problem Set 5.6 - Page 224: 84

$\dfrac{1}{5x^{3}}$

Using the laws of exponents, the given expression, $ \left( \dfrac{35x^{2}}{7x^{-1}} \right)^{-1} ,$ simplifies to \begin{array}{l}\require{cancel} \left( (35\div7)x^{2-(-1)} \right)^{-1} \\\\= \left( 5x^{2+1} \right)^{-1} \\\\= \left( 5x^{3} \right)^{-1} \\\\= \dfrac{1}{\left( 5x^{3} \right)^{1}} \\\\= \dfrac{1}{5x^{3}} .\end{array}