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

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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.
The simplified form of the expression is $$ 2^{A - N - 2} \cdot 3^{A - N + 1} $$.

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