Simplify using the product and quotient properties of radicals
cube root of 4X^4/25Y^6
$\displaystyle \displaystyle \sqrt[3]{\frac{4x^4}{25y^6}} = \frac{\sqrt[3]{4x^4}}{\sqrt[3]{25y^6}}$
$\displaystyle \displaystyle = \frac{\sqrt[3]{4}\sqrt[3]{x^4}}{\sqrt[3]{25}\sqrt[3]{y^6}}$
$\displaystyle \displaystyle = \frac{\sqrt[3]{4}\sqrt[3]{x^4}}{\sqrt[3]{25}\,y^2}$
$\displaystyle \displaystyle = \frac{\sqrt[3]{5}\sqrt[3]{4}\sqrt[3]{x^4}}{\sqrt[3]{5}\sqrt[3]{5^2}\,y^2}$
$\displaystyle \displaystyle = \frac{\sqrt[3]{5\cdot 4\cdot x^4}}{\sqrt[3]{5^3}\,y^2}$
$\displaystyle \displaystyle = \frac{\sqrt[3]{20x^4}}{5y^2}$.
You are asked to calculate partially the cube root of a term:
$\displaystyle \sqrt[3]{\dfrac{4x^4}{25y^6}} = \sqrt[3]{\dfrac{4x^4 \cdot 2 \cdot 5}{25y^6 \cdot 2 \cdot 5}} = $
Now re-arrange:
$\displaystyle \sqrt[3]{\dfrac{2^3 x^3 \cdot 5x}{5^3 y^6 \cdot 2 }} = $
Can you take it from here?
Hello, victorfk06!
$\displaystyle \text{Simplify using the product and quotient properties of radicals:}$
. . $\displaystyle \sqrt[3]{\dfrac{4x^4}{25y^6}}$
$\displaystyle \displaystyle \text{We have: }\;\sqrt[3]{\frac{4\cdot x^3 \cdot x}{5^2\cdot y^6}} $
$\displaystyle \displaystyle \text{Under the radical multiply by }\frac{5}{5}\!:\;\;\sqrt[3]{\frac{5}{5}\cdot \frac{4\cdot x^3 \cdot x}{5^2\cdot y^6}} $
$\displaystyle \displaystyle \text{We have: }\; \sqrt[3]{\frac{20\cdot x^3 \cdot x}{5^3\cdot y^6}} \;=\;\frac{\sqrt[3]{20}\cdot\sqrt[3]{x^3}\cdot\sqrt[3]{x}} {\sqrt[3]{5^3}\cdot\sqrt[3]{y^6}}$
. . . . . . . . $\displaystyle \displaystyle =\;\frac{\sqrt[3]{20}\cdot x \cdot \sqrt[3]{x}}{5\cdot y^2} \;=\;\frac{x\sqrt[3]{20x}}{5y^2}$