Simplify the expression: cube root of (2x^4 y^7) times cube root of (4x^2 y^2)

1. Combine under a single radical: Since both terms are cube roots, we can multiply the expressions inside the radicals together under a single cube root. This uses the property $\sqrt[n]{a} \cdot \sqrt[n]{b} = \sqrt[n]{ab}$.
$$ \sqrt[3]{2x^4 y^7} \cdot \sqrt[3]{4x^2 y^2} = \sqrt[3]{(2x^4 y^7)(4x^2 y^2)} $$
2. Multiply the terms inside the radical: Multiply the coefficients and add the exponents of the like variables.
$$ \sqrt[3]{(2 \cdot 4) x^{4+2} y^{7+2}} = \sqrt[3]{8x^6 y^9} $$
3. Simplify the radical: Find the cube root of each factor. We are looking for factors that are perfect cubes.
- The cube root of $8$ is $2$ since $2^3 = 8$.
- The cube root of $x^6$ is $x^2$ since $(x^2)^3 = x^6$.
- The cube root of $y^9$ is $y^3$ since $(y^3)^3 = y^9$.
$$ \sqrt[3]{8x^6 y^9} = \sqrt[3]{8} \cdot \sqrt[3]{x^6} \cdot \sqrt[3]{y^9} = 2x^2 y^3 $$