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

Let's break down the expression \( \frac{3^{x+1} \cdot 18^{x}}{6^{x+1} \cdot 9^{x+1}} \) into its prime factors for simplification. 1. We know that: - \( 18 = 2 \cdot 3^2 \), so \( 18^x = 2^x \cdot 3^{2x} \). - \( 6 = 2 \cdot 3 \), so \( 6^{x+1} = 2^{x+1} \cdot 3^{x+1} \). - \( 9 = 3^2 \), so \( 9^{x+1} = 3^{2(x+1)} = 3^{2x+2} \). 2. Substitute these back into the original expression: \[ \frac{3^{x+1} \cdot (2^x \cdot 3^{2x})}{(2^{x+1} \cdot 3^{x+1}) \cdot 3^{2x + 2}} \] This becomes: \[ = \frac{3^{x+1 + 2x} \cdot 2^x}{2^{x+1} \cdot 3^{x+1 + 2x + 2}} \] 3. Simplifying the powers of \( 3 \) and \( 2 \): The numerator has \( 3^{3x + 1} \cdot 2^x \), and the denominator has \( 2^{x+1} \cdot 3^{3x + 3} \). So the expression can be rewritten as: \[ = \frac{3^{3x+1} \cdot 2^x}{2^{x+1} \cdot 3^{3x + 3}} \] From here, we can further simplify: \[ = \frac{3^{3x + 1 - (3x + 3)} \cdot 2^{x - (x + 1)}}{1} = \frac{3^{-2} \cdot 2^{-1}}{1} = \frac{1}{9 \cdot 2} = \frac{1}{18} \] So the final result of simplifying the original expression is: \[ \boxed{\frac{1}{18}} \]