Answer
$$\frac{3\left(x+4\right)}{x+3}$$
Work Step by Step
Using the rules of exponents, we find:
$$\frac{3\left(x-3\right)\left(x^2-16\right)}{\left(x-4\right)\left(x^2-9\right)}\\ \frac{3\left(x-3\right)\left(x+4\right)\left(x-4\right)}{\left(x-4\right)\left(x+3\right)\left(x-3\right)}\\ \frac{3\left(x+4\right)}{x+3}$$
