Which of the following is equivalent to \( \frac{\sqrt[3]{32 x^{3} y^{6}}}{\sqrt[3]{2 x^{9} y^{2}}} \) where \( x \geq 0 \) and \( y \geq 0 \) ?

To simplify \( \frac{\sqrt[3]{32 x^{3} y^{6}}}{\sqrt[3]{2 x^{9} y^{2}}} \), we can combine the cube roots as follows: \[ \sqrt[3]{\frac{32 x^{3} y^{6}}{2 x^{9} y^{2}}} \] Now, simplifying the expression inside the cube root: 1. **Numerator:** \( 32 = 16 \times 2 \) which equals \( 2^5 \), hence we can express \( 32 \) as \( 2^5 \). 2. **Denominator:** \( 2 \) remains the same. When dividing the coefficients, we have: \[ \frac{32}{2} = \frac{2^5}{2^1} = 2^{5-1} = 2^4 = 16 \] For the variables \( x \) and \( y \): \[ \frac{x^{3}}{x^{9}} = x^{3-9} = x^{-6} = \frac{1}{x^6} \] \[ \frac{y^{6}}{y^{2}} = y^{6-2} = y^{4} \] Combining all these results inside the cube root: \[ \sqrt[3]{16 \cdot \frac{y^{4}}{x^{6}}} \] Calculating the terms: \[ \sqrt[3]{16} = 2^{4/3} = 2^{1+\frac{1}{3}} = 2 \cdot \sqrt[3]{2} \] Thus, we can write: \[ \sqrt[3]{\frac{16y^{4}}{x^{6}}} = \frac{2\sqrt[3]{2}y^{4/3}}{x^{2}} \] Putting the pieces together, the final simplified expression is: \[ \frac{2\sqrt[3]{2} y^{\frac{4}{3}}}{x^{2}} \] This expression represents the solution to the original problem, given that \( x \geq 0 \) and \( y \geq 0 \).