Answer
$\frac{x-1}{x+2}, x\ne 3, x\ne -1, x\ne 2, x\ne -2$
Work Step by Step
Factors of $x^{2}-5x+6$ is $(x-3)(x-2)$
Factors of $x^{2}-2x-3$ is $(x-3)(x+1)$
So $x^{2}-5x+6$ can be written as $(x-3)(x-2)$ and $x^{2}-2x-3$ can be written as $(x-3)(x+1)$
$(x^{2}-1)$ and $(x^{2}-4)$ are the difference of squares. The factors are $(x+1)(x-1)$ and $(x+2)(x-2)$
$=\frac{(x-3)(x-2)}{(x-3)(x+1)}\times \frac{(x+1)(x-1)}{(x+2)(x-2)}$
The denominator is $(x-3)(x+1)(x+2)(x-2)$. So $x\ne 3, x\ne -1, x\ne -2, x\ne 2$ otherwise that make the denominator zero.
$=\frac{(x-3)(x-2)}{(x-3)(x+1)}\times \frac{(x+1)(x-1)}{(x+2)(x-2)}$ ;$x\ne 3, x\ne -1, x\ne -2, x\ne 2$
Divide out common factors.
$=\frac{x-1}{x+2}, x\ne 3, -1, 2, -2$
