(b) \( \frac{25^{n} \cdot 15^{n-1} \cdot 3^{n}}{125^{n} \cdot 9^{n-1}} \)

To simplify the expression \( \frac{25^{n} \cdot 15^{n-1} \cdot 3^{n}}{125^{n} \cdot 9^{n-1}} \), we can rewrite each of the bases in terms of their prime factors: - \( 25 = 5^2 \) so \( 25^n = (5^2)^n = 5^{2n} \) - \( 15 = 3 \cdot 5 \) so \( 15^{n-1} = (3 \cdot 5)^{n-1} = 3^{n-1} \cdot 5^{n-1} \) - \( 3^{n} \) remains as is. - \( 125 = 5^3 \) so \( 125^n = (5^3)^n = 5^{3n} \) - \( 9 = 3^2 \) so \( 9^{n-1} = (3^2)^{n-1} = 3^{2(n-1)} = 3^{2n-2} \) Now substituting these into the expression, we have: \[ \frac{5^{2n} \cdot (3^{n-1} \cdot 5^{n-1}) \cdot 3^{n}}{5^{3n} \cdot 3^{2n-2}} \] This simplifies to: \[ \frac{5^{2n} \cdot 3^{n-1} \cdot 5^{n-1} \cdot 3^{n}}{5^{3n} \cdot 3^{2n-2}} = \frac{5^{2n+n-1} \cdot 3^{n+n-1}}{5^{3n} \cdot 3^{2n-2}} = \frac{5^{3n-1} \cdot 3^{2n-1}}{5^{3n} \cdot 3^{2n-2}} \] Combining the powers of the bases gives us: \[ = \frac{5^{3n-1-3n} \cdot 3^{2n-1-(2n-2)}} = \frac{5^{-1} \cdot 3^{1}}{1} = \frac{3}{5} \] Thus, the simplified expression is: \[ \frac{3}{5} \]