The problem asks to simplify the expression $\sqrt[3]{24x^6y^{11}}$. We assume all variables are nonnegative real numbers.

1. Problem Description

The problem asks to simplify the expression $\sqrt[3]{24x^6y^{11}}$. We assume all variables are nonnegative real numbers.

2. Solution Steps

We need to simplify $\sqrt[3]{24x^6y^{11}}$.

First, we can rewrite 24 as its prime factorization: $24 = 2^3 \cdot 3$.

So, the expression becomes $\sqrt[3]{2^3 \cdot 3 \cdot x^6 \cdot y^{11}}$.

We can separate the expression as $\sqrt[3]{2^3} \cdot \sqrt[3]{3} \cdot \sqrt[3]{x^6} \cdot \sqrt[3]{y^{11}}$.

We know that $\sqrt[3]{a^3} = a$ and $\sqrt[3]{x^6} = x^{6/3} = x^2$.

Therefore, $\sqrt[3]{2^3} = 2$ and $\sqrt[3]{x^6} = x^2$.

For $\sqrt[3]{y^{11}}$, we can rewrite $y^{11}$ as $y^9 \cdot y^2$, where 9 is the largest multiple of 3 less than or equal to

1

1. Then, $\sqrt[3]{y^{11}} = \sqrt[3]{y^9 \cdot y^2} = \sqrt[3]{y^9} \cdot \sqrt[3]{y^2} = y^{9/3} \cdot \sqrt[3]{y^2} = y^3 \sqrt[3]{y^2}$.

Substituting these back into the expression gives:

$2 \cdot \sqrt[3]{3} \cdot x^2 \cdot y^3 \sqrt[3]{y^2} = 2 x^2 y^3 \sqrt[3]{3y^2}$.

3. Final Answer

$2x^2y^3\sqrt[3]{3y^2}$