Identity

$\begin{aligned} \frac {x^2-x-6}{x^2+6x+8} \cdot \frac {x+4}{x-3} & = \frac {(x+2)(x-3)}{(x+2)(x+4)}\cdot \frac {x+4}{x-3} = \frac {\cancel{(x+2)}\cancel{(x-3)}}{\cancel{(x+2)} \cancel{(x+4)}}\cdot \frac {\cancel{x+4}}{\cancel{x-3}} = \boxed{1} \end{aligned}$

We can simplify the denominator of the first fraction with the numerator of the second fraction, resulting in x + 2 . Next we divide the numerator of the first fraction by x + 2 to get x - 3 . (x - 3)/(x - 3) = $\boxed{1}$ .