(b) \( \frac{2^{2 n} \cdot 2^{n} \cdot 2^{n}-8^{n}}{2^{5} \cdot 2^{3 n}} \)

To simplify the expression \[ (b) \frac{2^{2n} \cdot 2^{n} \cdot 2^{n} - 8^{n}}{2^{5} \cdot 2^{3n}}, \] we start by recognizing that \( 8^{n} \) can be rewritten using the base of 2, since \( 8 = 2^{3} \). Thus, \[ 8^{n} = (2^{3})^{n} = 2^{3n}. \] Now substituting this into the expression, we have: \[ \frac{2^{2n} \cdot 2^{n} \cdot 2^{n} - 2^{3n}}{2^{5} \cdot 2^{3n}}. \] Now simplify the numerator. We can combine the powers of 2: \[ 2^{2n} \cdot 2^{n} \cdot 2^{n} = 2^{2n + n + n} = 2^{4n}. \] Thus, the numerator simplifies to: \[ 2^{4n} - 2^{3n}. \] We can factor out \( 2^{3n} \) from the numerator: \[ 2^{3n}(2^{n} - 1). \] Now putting it all together, we can rewrite the whole expression: \[ \frac{2^{3n}(2^{n} - 1)}{2^{5} \cdot 2^{3n}}. \] The \( 2^{3n} \) in the numerator and denominator cancels out: \[ \frac{2^{n} - 1}{2^{5}}. \] This gives us our simplified expression: \[ \frac{2^{n} - 1}{32}. \] Hence, the final answer is: \[ (b) \frac{2^{n} - 1}{32}. \]