a) \frac{18^{n+3}}{3^{2n+5} \cdot 2^{n-2}}

\frac{18^{n+3}}{3^{2n+5} \cdot 2^{n-2}} = \frac{(2 \cdot 3^2)^{n+3}}{3^{2n+5} \cdot 2^{n-2}} = \frac{2^{n+3} \cdot 3^{2(n+3)}}{3^{2n+5} \cdot 2^{n-2}} = \frac{2^{n+3} \cdot 3^{2n+6}}{3^{2n+5} \cdot 2^{n-2}} = 2^{(n+3)-(n-2)} \cdot 3^{(2n+6)-(2n+5)} = 2^{n+3-n+2} \cdot 3^{2n+6-2n-5} = 2^5 \cdot 3^1 = 32 \cdot 3 = 96