1. **State the problem:** Simplify the expression $$\frac{x+5}{9x+27} \cdot \frac{x^2 - 9}{x^2 + 2x - 15}$$. 2. **Factor all polynomials:** - Factor the denominator $9x+27$ as $9(x+3)$. - Factor the numerator $x^2 - 9$ as a difference of squares: $(x-3)(x+3)$. - Factor the denominator $x^2 + 2x - 15$ by finding two numbers that multiply to $-15$ and add to $2$: $(x+5)(x-3)$. 3. **Rewrite the expression with factored forms:** $$\frac{x+5}{9(x+3)} \cdot \frac{(x-3)(x+3)}{(x+5)(x-3)}$$ 4. **Cancel common factors:** - Cancel $(x+5)$ from numerator and denominator. - Cancel $(x-3)$ from numerator and denominator. - Cancel $(x+3)$ from numerator and denominator. Intermediate step showing cancellation: $$\frac{\cancel{x+5}}{9\cancel{(x+3)}} \cdot \frac{\cancel{(x-3)}\cancel{(x+3)}}{\cancel{(x+5)}\cancel{(x-3)}} = \frac{1}{9} \cdot 1 = \frac{1}{9}$$ 5. **Final simplified expression:** $$\boxed{\frac{1}{9}}$$ This means the original expression simplifies neatly to $\frac{1}{9}$ after factoring and canceling common terms.