12. \( \frac{x^{2}-7 x-8}{x^{2}-64} \cdot \frac{x+8}{x^{2}+4 x+3} \)

To simplify the expression \( \frac{x^{2}-7 x-8}{x^{2}-64} \cdot \frac{x+8}{x^{2}+4 x+3} \), we can start by factoring each part. 1. **Factor the numerator and denominator**: - \( x^2 - 7x - 8 = (x - 8)(x + 1) \) - \( x^2 - 64 = (x - 8)(x + 8) \) (difference of squares) - \( x + 8 \) remains as is. - \( x^2 + 4x + 3 = (x + 1)(x + 3) \) Putting this all together, we rewrite the expression: \[ \frac{(x - 8)(x + 1)}{(x - 8)(x + 8)} \cdot \frac{x + 8}{(x + 1)(x + 3)} \] 2. **Cancel common factors**: - The \( (x - 8) \) in the numerator and denominator cancels. - The \( (x + 8) \) in the numerator of the second fraction cancels with the \( (x + 8) \) in the denominator of the first fraction. - The \( (x + 1) \) in the numerator cancels with the one in the denominator. Now the expression simplifies to: \[ \frac{1}{x + 3} \] Thus, the final simplified expression is: \[ \frac{1}{x + 3} \] Where \( x \neq -3 \) to avoid division by zero!