Practice simplifying higher-order roots whose radicands are complex fractions by first reducing the fraction inside the radical, then applying the Quotient Property of $n$th Roots.

ⓐ $\sqrt[3]{\frac{54x^7y^5}{250x^2y^2}} = \frac{3xy\sqrt[3]{x^2}}{5}$: Simplify the fraction inside the radicand: cancel the shared factor of $2$ from the coefficients ($\frac{54}{250} = \frac{27}{125}$) and apply the Quotient Property for Exponents to the variables ($\frac{x^7}{x^2} = x^5$ and $\frac{y^5}{y^2} = y^3$). The expression becomes $\sqrt[3]{\frac{27x^5y^3}{125}}$. Rewrite using the Quotient Property: $\frac{\sqrt[3]{27x^5y^3}}{\sqrt[3]{125}}$. The denominator evaluates cleanly to $5$. Simplify the numerator by extracting perfect cube factors: $\sqrt[3]{27x^3y^3} \cdot \sqrt[3]{x^2} = 3xy\sqrt[3]{x^2}$. The simplified form is $\frac{3xy\sqrt[3]{x^2}}{5}$.

ⓑ $\sqrt[4]{\frac{32a^9b^7}{162a^3b^3}} = \frac{2/ab/\sqrt[4]{a^2}}{3}$: Simplify the fraction inside the radicand: cancel the shared factor of $2$ ($\frac{32}{162} = \frac{16}{81}$) and simplify the variables ($\frac{a^9}{a^3} = a^6$ and $\frac{b^7}{b^3} = b^4$). The expression becomes $\sqrt[4]{\frac{16a^6b^4}{81}}$. Rewrite using the Quotient Property: $\frac{\sqrt[4]{16a^6b^4}}{\sqrt[4]{81}}$. The denominator evaluates to $3$. Simplify the numerator by extracting perfect fourth power factors: $\sqrt[4]{16a^4b^4} \cdot \sqrt[4]{a^2}$. Because the index is even, taking the principal fourth root of $16a^4b^4$ requires absolute value signs: $2/ab/$. The simplified form is $\frac{2/ab/\sqrt[4]{a^2}}{3}$.