\( \frac { 6 ^ { A + 3 } \cdot 2 ^ { N - 1 } } { 12 ^ { N + 2 } } \)

To simplify the expression \[ \frac { 6 ^ { A + 3 } \cdot 2 ^ { N - 1 } } { 12 ^ { N + 2 } } \] we will first express all terms in the numerator and denominator using the same base. 1. **Rewrite the bases:** - \( 6 \) can be expressed as \( 2 \cdot 3 \). - \( 12 \) can be expressed as \( 2^2 \cdot 3 \). Thus, we can rewrite the expression as: \[ \frac { (2 \cdot 3)^{A + 3} \cdot 2^{N - 1} } { (2^2 \cdot 3)^{N + 2} } \] 2. **Expand the powers:** - The numerator becomes: \[ (2^{A + 3} \cdot 3^{A + 3}) \cdot 2^{N - 1} = 2^{A + 3 + N - 1} \cdot 3^{A + 3} = 2^{A + N + 2} \cdot 3^{A + 3} \] - The denominator becomes: \[ (2^2)^{N + 2} \cdot 3^{N + 2} = 2^{2(N + 2)} \cdot 3^{N + 2} = 2^{2N + 4} \cdot 3^{N + 2} \] 3. **Combine the expression:** Now we can rewrite the entire expression as: \[ \frac { 2^{A + N + 2} \cdot 3^{A + 3} } { 2^{2N + 4} \cdot 3^{N + 2} } \] 4. **Simplify the expression:** - For the base \( 2 \): \[ 2^{A + N + 2 - (2N + 4)} = 2^{A + N + 2 - 2N - 4} = 2^{A - N - 2} \] - For the base \( 3 \): \[ 3^{A + 3 - (N + 2)} = 3^{A + 3 - N - 2} = 3^{A - N + 1} \] 5. **Final expression:** Combining these results, we have: \[ \frac { 6^{A + 3} \cdot 2^{N - 1} } { 12^{N + 2} } = 2^{A - N - 2} \cdot 3^{A - N + 1} \] This is the simplified form of the original expression.