\( \begin{array}{ll}\text { (a) } \frac{8^{x+1} \cdot 2^{2 x-1}}{32^{x}} & \text { (b) } \frac{3^{n} \cdot 27^{n+1}}{9^{2 n+3}}\end{array} \)

1. For the expression \[ \frac{8^{x+1} \cdot 2^{2x-1}}{32^x}, \] start by writing all bases in terms of \(2\). - Note that \(8=2^3\), so \[ 8^{x+1}=(2^3)^{x+1}=2^{3(x+1)}. \] - Also, \(32=2^5\), so \[ 32^x=(2^5)^x=2^{5x}. \] Substitute into the original expression: \[ \frac{2^{3(x+1)} \cdot 2^{2x-1}}{2^{5x}}. \] Combine the exponents in the numerator: \[ 2^{3(x+1)} \cdot 2^{2x-1} = 2^{3x+3+2x-1} = 2^{5x+2}. \] Now, the expression becomes: \[ \frac{2^{5x+2}}{2^{5x}}=2^{(5x+2)-5x}=2^2=4. \] 2. For the expression \[ \frac{3^n \cdot 27^{n+1}}{9^{2n+3}}, \] write all bases in terms of \(3\). - Since \(27=3^3\), we have \[ 27^{n+1}=(3^3)^{n+1}=3^{3(n+1)}=3^{3n+3}. \] - And since \(9=3^2\), \[ 9^{2n+3}=(3^2)^{2n+3}=3^{2(2n+3)}=3^{4n+6}. \] Substitute these into the original expression: \[ \frac{3^n \cdot 3^{3n+3}}{3^{4n+6}}. \] Combine the exponents in the numerator: \[ 3^n \cdot 3^{3n+3} = 3^{n+3n+3} = 3^{4n+3}. \] Now, the expression simplifies to: \[ \frac{3^{4n+3}}{3^{4n+6}} = 3^{(4n+3)-(4n+6)} = 3^{-3}=\frac{1}{3^3}=\frac{1}{27}. \]