## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$-\dfrac{9x^3}{(x-3)^2}$
We apply Method 1 (A51), to treat the numerator and denominator of the complex rational expression separately. So by factoring both the denominator and numerator of each rational function and cancelling the common factors, we have $$\frac{\frac{3+x}{3-x}}{\frac{x^2-9}{9x^3}}=\frac{\frac{3+x}{3-x}}{\frac{(x-3)(x+3)}{9x^3}}= \frac{3+x}{3-x}\frac{9x^3}{(x-3)(x+3)}\\ =-\frac{3+x}{x-3}\frac{9x^3}{(x-3)(x+3)} =-\frac{9x^3}{(x-3)^2} .$$