Simplify \( \frac{3^{x+3} \cdot 18^{x}}{6^{x+1} \cdot 9^{x+1}} \)

To simplify the expression \( \frac{3^{x+3} \cdot 18^{x}}{6^{x+1} \cdot 9^{x+1}} \), we first need to express all terms in bases of 3 and 2: 1. Rewrite the bases: \[ 18 = 2 \cdot 3^2 \quad \Rightarrow \quad 18^x = 2^x \cdot 3^{2x} \] \[ 6 = 2 \cdot 3 \quad \Rightarrow \quad 6^{x+1} = 2^{x+1} \cdot 3^{x+1} \] \[ 9 = 3^2 \quad \Rightarrow \quad 9^{x+1} = 3^{2(x+1)} = 3^{2x+2} \] 2. Substitute these values back into the original expression: \[ = \frac{3^{x+3} \cdot (2^x \cdot 3^{2x})}{(2^{x+1} \cdot 3^{x+1}) \cdot (3^{2x+2})} \] \[ = \frac{3^{x+3} \cdot 2^x \cdot 3^{2x}}{2^{x+1} \cdot 3^{x+1} \cdot 3^{2x+2}} \] \[ = \frac{3^{x+3 + 2x} \cdot 2^x}{2^{x+1} \cdot 3^{(x+1) + (2x+2)}} \] 3. Combine the exponents: \[ = \frac{3^{3x + 3} \cdot 2^x}{2^{x+1} \cdot 3^{3x + 3}} \] 4. Now, cancel \( 3^{3x + 3} \): \[ = \frac{2^x}{2^{x+1}} \] 5. Simplify the powers of 2: \[ = \frac{1}{2^{x+1 - x}} = \frac{1}{2^1} = \frac{1}{2} \] Thus, the simplified expression is: \[ \frac{1}{2} \]