1Cademy - Simplifying $$\sqrt[3]{40x^4y^5}$$ and $$\sqrt[4]{48x^4y^7}$$

Learn Before

Simplifying $\sqrt[3]{24x^7}$ and $\sqrt[4]{80y^{14}}$

Example

Simplifying $\sqrt[3]{40x^4y^5}$ and $\sqrt[4]{48x^4y^7}$

Apply the Product Property of $n$ th Roots to simplify two higher-order radical expressions whose radicands contain a non-perfect-power numerical coefficient and two variables. Factor the numerical and variable parts separately to extract the largest perfect $n$ th power factor.

ⓐ $\sqrt[3]{40x^4y^5} = 2xy\sqrt[3]{5xy^2}$ : For the index $3$ , factor the radicand into perfect cube factors: $40 = 8 \cdot 5$ , $x^4 = x^3 \cdot x$ , and $y^5 = y^3 \cdot y^2$ . Rewrite as $\sqrt[3]{8x^3y^3 \cdot 5xy^2}$ . Separate using the Product Property: $\sqrt[3]{(2xy)^3} \cdot \sqrt[3]{5xy^2}$ . Simplify the perfect cube to obtain $2xy\sqrt[3]{5xy^2}$ . Since the index is odd, no absolute value is needed.

ⓑ $\sqrt[4]{48x^4y^7} = 2|xy|\sqrt[4]{3y^3}$ : For the index $4$ , factor into perfect fourth power factors: $48 = 16 \cdot 3$ , $x^4$ is already a perfect fourth power, and $y^7 = y^4 \cdot y^3$ . Rewrite as $\sqrt[4]{16x^4y^4 \cdot 3y^3}$ . Separate: $\sqrt[4]{(2xy)^4} \cdot \sqrt[4]{3y^3}$ . Because the index $4$ is even, variables extracted from the radical require absolute value signs: $\sqrt[4]{(2xy)^4} = 2|xy|$ . The simplified form is $2|xy|\sqrt[4]{3y^3}$ .