1. **State the problem:** Simplify the expression $$\frac{27^{n+2} - 6 \cdot 3^{3n+3}}{3^n \cdot 9^{n+2}}$$. 2. **Rewrite bases as powers of 3:** - Note that $27 = 3^3$ and $9 = 3^2$. - So, rewrite each term: $$27^{n+2} = (3^3)^{n+2} = 3^{3(n+2)} = 3^{3n+6}$$ $$3^{3n+3}$$ stays as is. $$9^{n+2} = (3^2)^{n+2} = 3^{2(n+2)} = 3^{2n+4}$$ 3. **Substitute back into the expression:** $$\frac{3^{3n+6} - 6 \cdot 3^{3n+3}}{3^n \cdot 3^{2n+4}} = \frac{3^{3n+6} - 6 \cdot 3^{3n+3}}{3^{n + 2n + 4}} = \frac{3^{3n+6} - 6 \cdot 3^{3n+3}}{3^{3n+4}}$$ 4. **Factor numerator:** Factor out $3^{3n+3}$ from the numerator: $$3^{3n+3}(3^3 - 6) = 3^{3n+3}(27 - 6) = 3^{3n+3} \cdot 21$$ 5. **Rewrite the expression:** $$\frac{3^{3n+3} \cdot 21}{3^{3n+4}}$$ 6. **Simplify powers of 3:** $$\frac{3^{3n+3} \cdot 21}{3^{3n+4}} = 21 \cdot \frac{3^{3n+3}}{3^{3n+4}} = 21 \cdot 3^{\cancel{3n+3} - \cancel{3n+4}} = 21 \cdot 3^{-1}$$ 7. **Simplify further:** $$21 \cdot 3^{-1} = \frac{21}{3} = 7$$ **Final answer:** $$7$$