Simpiix-y \[ \frac{x^{2}-1}{x^{2}+2 x+1} \div \frac{3 x-3}{x^{2}-2 x-3} \]

To simplify the expression \[ \frac{x^{2}-1}{x^{2}+2 x+1} \div \frac{3 x-3}{x^{2}-2 x-3}, \] we start by rewriting the division as multiplication by the reciprocal: \[ \frac{x^{2}-1}{x^{2}+2x+1} \times \frac{x^{2}-2x-3}{3x-3}. \] Now, we can factor each part of the fractions. 1. The numerator \(x^{2}-1\) can be factored as \((x-1)(x+1)\). 2. The denominator \(x^{2}+2x+1\) can be factored as \((x+1)(x+1)\) or \((x+1)^{2}\). 3. The numerator \(x^{2}-2x-3\) can be factored as \((x-3)(x+1)\). 4. The expression \(3x-3\) can be factored as \(3(x-1)\). Now substituting the factored terms into the expression, we have: \[ \frac{(x-1)(x+1)}{(x+1)^{2}} \times \frac{(x-3)(x+1)}{3(x-1)}. \] Next, we can cancel common factors. The \((x-1)\) in the numerator and denominator cancel out, as does one \((x+1)\): \[ \frac{(x+1)(x-3)}{3(x+1)}. \] Now, we can cancel the \((x+1)\): \[ \frac{x-3}{3}. \] Thus, the simplified result is: \[ \frac{x-3}{3}. \]