Answer
$\frac{x+3}{x-2} ; x\ne2,-2,3,-3$
Work Step by Step
The factors of $(x^{2}+5x+6 )$ is $(x+3)$ and $(x+2)$.
The expression $(x^{2}+x-6 )$'s factors are $(x+3)$ and $(x-2)$.
$(x^{2}-9)$ is the difference of squares, so the factors are $(x-3)$ and $(x+3)$.
$(x^{2}-x-6 )$'s factors are $(x-3)$ and $(x+2)$.
The given expression can be written as
$= \frac{(x+3)(x+2)}{(x+3)(x-2)} . \frac{(x+3)(x-3)}{(x-3)(x+2)}$
The denominators are $(x+3)(x-2)(x-3)(x+2)$. So the numbers 2,-2,3,-3 must be excluded, otherwise they will make the denominator zero.
$= \frac{(x+3)(x+2)}{(x+3)(x-2)} . \frac{(x+3)(x-3)}{(x-3)(x+2)} ; x\ne 2,-2,3,-3;$
Divide out common factors from numerator and denominator. The result is
$=\frac{x+3}{x-2} ; x\ne 2,-2,3,-3;$
