D. 12. \( [2002 / 1] \) Evaluate \( \frac{(81)^{3 / 4}-(27)^{1 / 3}}{3 \times 2^{3}} \) A. 1. B. \( \frac{1}{8} \) C. 3. D. \( \frac{1}{3} \cdot[2005 / 7] \)

To evaluate \( \frac{(81)^{3 / 4}-(27)^{1 / 3}}{3 \times 2^{3}} \), we start solving each part step by step. First, we simplify \( (81)^{3/4} \): \[ 81 = 3^4 \implies (81)^{3/4} = (3^4)^{3/4} = 3^{4 \cdot (3/4)} = 3^3 = 27. \] Next, we simplify \( (27)^{1/3} \): \[ 27 = 3^3 \implies (27)^{1/3} = (3^3)^{1/3} = 3^{3 \cdot (1/3)} = 3. \] Now substitute these values back into the expression: \[ \frac{27 - 3}{3 \times 2^3}. \] Calculate the numerator: \[ 27 - 3 = 24. \] And for the denominator, calculate \( 2^3 \): \[ 2^3 = 8 \implies 3 \times 2^{3} = 3 \times 8 = 24. \] Now we have: \[ \frac{24}{24} = 1. \] Thus, the final answer is \( \boxed{1} \).