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

We start with \[ \frac{x^{2}-1}{x^{2}+2 x+1} \div \frac{3 x-3}{x^{2}-2 x-3}. \] **Step 1. Factor the expressions.** - Factor the numerator of the first fraction: \[ x^2-1 = (x-1)(x+1). \] - Factor the denominator of the first fraction: \[ x^2+2x+1 = (x+1)^2. \] - Factor the numerator of the second fraction: \[ 3x-3 = 3(x-1). \] - Factor the denominator of the second fraction: \[ x^2-2x-3 = (x-3)(x+1). \] **Step 2. Rewrite the expression with factorizations.** The original expression becomes \[ \frac{(x-1)(x+1)}{(x+1)^2} \div \frac{3(x-1)}{(x-3)(x+1)}. \] **Step 3. Change division to multiplication by the reciprocal.** \[ \frac{(x-1)(x+1)}{(x+1)^2} \times \frac{(x-3)(x+1)}{3(x-1)}. \] **Step 4. Cancel common factors.** - Cancel the common factor \(x-1\) from the numerator and denominator: \[ \frac{(x-1)(x+1)}{(x+1)^2} \times \frac{(x-3)(x+1)}{3(x-1)} = \frac{(x+1)}{(x+1)^2} \times \frac{(x-3)(x+1)}{3}. \] - Cancel one factor of \(x+1\) from the numerator of the first fraction and the denominator: \[ \frac{1}{x+1} \times \frac{(x-3)(x+1)}{3}. \] - Cancel the remaining common factor \(x+1\): \[ \frac{x-3}{3}. \] **Final Answer:** \[ \frac{x-3}{3} \] *Note: Remember to exclude the values that make any original denominator zero, namely \(x\neq -1\), \(x\neq 1\), and \(x\neq 3\).*