We are asked to simplify the expression $\frac{2^3 \cdot 4^{-1} \cdot 8^2}{16^{-1} \cdot 32^{\frac{1}{5}}}$.

1. Problem Description

We are asked to simplify the expression $\frac{2^3 \cdot 4^{-1} \cdot 8^2}{16^{-1} \cdot 32^{\frac{1}{5}}}$.

2. Solution Steps

First, rewrite all numbers as powers of

2. $4 = 2^2$

$8 = 2^3$

$16 = 2^4$

$32 = 2^5$

Substitute these into the original expression:

$\frac{2^3 \cdot (2^2)^{-1} \cdot (2^3)^2}{(2^4)^{-1} \cdot (2^5)^{\frac{1}{5}}}$

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

$\frac{2^3 \cdot 2^{-2} \cdot 2^6}{2^{-4} \cdot 2^1}$

Using the rule $a^m \cdot a^n = a^{m+n}$, we simplify the numerator and denominator separately:

Numerator: $2^3 \cdot 2^{-2} \cdot 2^6 = 2^{3 + (-2) + 6} = 2^{7}$

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

Now, divide the numerator by the denominator using the rule $\frac{a^m}{a^n} = a^{m-n}$:

$\frac{2^7}{2^{-3}} = 2^{7 - (-3)} = 2^{7 + 3} = 2^{10}$

Finally, evaluate $2^{10}$.

$2^{10} = 1024$

3. Final Answer

1024