Simplify the expression $\sqrt[3]{192x^3y^5z^{10}}$.

1. Problem Description

Simplify the expression $\sqrt[3]{192x^3y^5z^{10}}$.

2. Solution Steps

First, we find the prime factorization of

1

9

2. $192 = 2^6 \cdot 3$.

Then we rewrite the given expression:

$\sqrt[3]{192x^3y^5z^{10}} = \sqrt[3]{2^6 \cdot 3 \cdot x^3 \cdot y^5 \cdot z^{10}}$.

We can rewrite $y^5$ as $y^3 \cdot y^2$ and $z^{10}$ as $z^9 \cdot z$.

$\sqrt[3]{2^6 \cdot 3 \cdot x^3 \cdot y^3 \cdot y^2 \cdot z^9 \cdot z} = \sqrt[3]{(2^2)^3 \cdot 3 \cdot x^3 \cdot y^3 \cdot y^2 \cdot (z^3)^3 \cdot z}$

Now we take out the perfect cubes from the cube root:

$\sqrt[3]{(2^2)^3 \cdot x^3 \cdot y^3 \cdot (z^3)^3 \cdot 3y^2z} = 2^2xyz^3\sqrt[3]{3y^2z} = 4xyz^3\sqrt[3]{3y^2z}$.

3. Final Answer

The simplified expression is $4xyz^3\sqrt[3]{3y^2z}$.