1. The problem is to simplify the expression $$\frac{x^2 - 9}{x + 3}$$. 2. Recognize that the numerator is a difference of squares: $$x^2 - 9 = (x - 3)(x + 3)$$. 3. Substitute this factorization back into the expression: $$\frac{(x - 3)(x + 3)}{x + 3}$$. 4. For $$x \neq -3$$, cancel the common factor $$x + 3$$ in numerator and denominator: $$\frac{(x - 3)\cancel{(x + 3)}}{\cancel{x + 3}} = x - 3$$. 5. The simplified expression is $$x - 3$$, with the restriction that $$x \neq -3$$ to avoid division by zero. 6. Therefore, the simplified form is $$x - 3$$, but note the domain excludes $$x = -3$$. Final answer: $$x - 3$$ for $$x \neq -3$$.