Simplify \( \frac{3^{2 n}-9^{n+1}}{72^{n}} \)

To simplify the expression \( \frac{3^{2 n}-9^{n+1}}{72^{n}} \), we start by recognizing that \( 9 = 3^2 \) and \( 72 = 2^3 \cdot 3^2 \). First, we rewrite \( 9^{n+1} \): \[ 9^{n+1} = (3^2)^{n+1} = 3^{2(n+1)} = 3^{2n + 2} \] Now, substituting this back into the expression gives: \[ \frac{3^{2n} - 3^{2n + 2}}{72^n} \] Next, factor out \( 3^{2n} \) from the numerator: \[ \frac{3^{2n}(1 - 3^2)}{72^n} = \frac{3^{2n}(1 - 9)}{72^n} = \frac{3^{2n}(-8)}{72^n} \] Now we compute \( 72^n \): \[ 72^n = (2^3 \cdot 3^2)^n = 2^{3n} \cdot 3^{2n} \] Substituting \( 72^n \) into the expression gives: \[ \frac{-8 \cdot 3^{2n}}{2^{3n} \cdot 3^{2n}} \] Now \( 3^{2n} \) in the numerator and denominator cancels out: \[ \frac{-8}{2^{3n}} \] Since \( 8 = 2^3 \), we can rewrite the fraction: \[ = \frac{-2^3}{2^{3n}} = -\frac{1}{2^{3n - 3}} = -\frac{1}{2^{3(n - 1)}} \] Thus, the simplified expression is: \[ \boxed{-\frac{1}{2^{3(n - 1)}}} \]