The problem is to evaluate the expression $(\frac{1}{32})^{\frac{-4}{3}}$.

1. Problem Description

The problem is to evaluate the expression $(\frac{1}{32})^{\frac{-4}{3}}$.

2. Solution Steps

First, we can rewrite the expression using the property $a^{-n} = \frac{1}{a^n}$:

$(\frac{1}{32})^{\frac{-4}{3}} = (\frac{32}{1})^{\frac{4}{3}} = (32)^{\frac{4}{3}}$

Next, we can rewrite 32 as $2^5$, so the expression becomes:

$(2^5)^{\frac{4}{3}}$

Using the property $(a^m)^n = a^{m \cdot n}$, we have:

$2^{5 \cdot \frac{4}{3}} = 2^{\frac{20}{3}}$

We can rewrite this as $2^{\frac{18}{3} + \frac{2}{3}} = 2^{6 + \frac{2}{3}} = 2^6 \cdot 2^{\frac{2}{3}}$

Since $2^6 = 64$, the expression becomes:

$64 \cdot 2^{\frac{2}{3}} = 64 \cdot (2^2)^{\frac{1}{3}} = 64 \cdot (4)^{\frac{1}{3}} = 64\sqrt[3]{4}$

3. Final Answer

$64\sqrt[3]{4}$