## College Algebra (10th Edition)

$-\displaystyle \frac{9x^{3}}{(x-3)^{2}}$
$...=\displaystyle \frac{3+x}{3-x}\div\frac{x^{2}-9}{9x^{3}}=\frac{3+x}{3-x}\cdot\frac{9x^{3}}{x^{2}-9}$ Factoring, $x^{2}-9=(x+3)(x-3)\quad$ (a difference of squares) $3-x=-(x-3)$ ... = $\displaystyle \frac{(x+3)\cdot 9x^{3}}{-(x-3)(x-3)(x+3)}$ ... cancel the following from both sides of the fraction line: $(x+3)$ ... = $\displaystyle \frac{9x^{3}}{-(x-3)(x-3)}$ = $-\displaystyle \frac{9x^{3}}{(x-3)^{2}}$