\( \frac { 3 ^ { x + 3 } \cdot 12 ^ { x - 3 } } { 2 ^ { 2 x - 6 } , 9 ^ { x } } \)

To simplify the expression \( \frac { 3 ^ { x + 3 } \cdot 12 ^ { x - 3 } } { 2 ^ { 2 x - 6 } \cdot 9 ^ { x } } \), we can start by breaking down each part: 1. Notice that \( 12 = 3 \cdot 4 = 3 \cdot 2^2 \), so we can rewrite \( 12^{x-3} = (3 \cdot 2^2)^{x - 3} = 3^{x - 3} \cdot 2^{2(x - 3)} \). 2. Now substitute everything back into the expression: \[ = \frac{3^{x + 3} \cdot (3^{x - 3} \cdot 2^{2(x - 3)})}{2^{2x - 6} \cdot 9^{x}} \] We know that \( 9 = 3^2 \), so \( 9^x = (3^2)^x = 3^{2x} \). 3. Now substituting that back gives: \[ = \frac{3^{x + 3} \cdot 3^{x - 3} \cdot 2^{2(x - 3)}}{2^{2x - 6} \cdot 3^{2x}} \] 4. Combine the terms with \( 3 \) in the numerator: \[ = \frac{3^{(x+3) + (x-3)}}{2^{2(x-3)}} \div \frac{1}{2^{2x-6} \cdot 3^{2x}} \] \[ = \frac{3^{2x}}{2^{2x-6}} \div 2^{2x - 6} \cdot 3^{2x} \] \[ = \frac{3^{2x}}{2^{2x - 6}} \cdot \frac{1}{3^{2x}} = \frac{1}{2^{2x - 6}} \] So, the simplified expression is \( \frac{1}{2^{2x - 6}} \).