## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$\dfrac{1}{3x}$
Simplify the radicand (expression inside the radicla sign) to obtain: $$\sqrt[3]{\frac{3xy^2}{81x^4y^2}}=\sqrt[3]{\frac{1}{27x^3}}==\sqrt[3]{\frac{1}{(3x)^3}}$$ Use the rule $\sqrt[n]{a^n}=a, a\gt0$ to obtain $$\sqrt[3]{\frac{1}{(3x)^3}}=\frac{1}{3x}$$