The problem asks to find the value of the expression $\frac{6^{\frac{1}{2}}9^{\frac{1}{4}}}{216^{\frac{1}{4}}}$.

1. Problem Description

The problem asks to find the value of the expression $\frac{6^{\frac{1}{2}}9^{\frac{1}{4}}}{216^{\frac{1}{4}}}$.

2. Solution Steps

First, we express each number in terms of its prime factors:

$6 = 2 \cdot 3$

$9 = 3^2$

$216 = 6^3 = (2 \cdot 3)^3 = 2^3 \cdot 3^3$

Now, substitute these into the given expression:

$\frac{6^{\frac{1}{2}}9^{\frac{1}{4}}}{216^{\frac{1}{4}}} = \frac{(2 \cdot 3)^{\frac{1}{2}} (3^2)^{\frac{1}{4}}}{(2^3 \cdot 3^3)^{\frac{1}{4}}}$

Using the power of a product rule $(ab)^n = a^n b^n$ and the power of a power rule $(a^m)^n = a^{mn}$, we get:

$\frac{2^{\frac{1}{2}} \cdot 3^{\frac{1}{2}} \cdot 3^{\frac{2}{4}}}{2^{\frac{3}{4}} \cdot 3^{\frac{3}{4}}} = \frac{2^{\frac{1}{2}} \cdot 3^{\frac{1}{2}} \cdot 3^{\frac{1}{2}}}{2^{\frac{3}{4}} \cdot 3^{\frac{3}{4}}}$

Now, we simplify the expression by combining the terms with the same base using the rule $\frac{a^m}{a^n} = a^{m-n}$:

$2^{\frac{1}{2} - \frac{3}{4}} \cdot 3^{\frac{1}{2} + \frac{1}{2} - \frac{3}{4}} = 2^{\frac{2}{4} - \frac{3}{4}} \cdot 3^{\frac{4}{4} - \frac{3}{4}} = 2^{-\frac{1}{4}} \cdot 3^{\frac{1}{4}}$

Rewriting this using the property $a^{-n} = \frac{1}{a^n}$, we have:

$\frac{3^{\frac{1}{4}}}{2^{\frac{1}{4}}} = (\frac{3}{2})^{\frac{1}{4}}$

3. Final Answer

$(\frac{3}{2})^{\frac{1}{4}}$