## Precalculus (6th Edition) Blitzer

$\displaystyle \frac{2(x^{2}+9)}{(x-3)(x+3)},\quad x\neq-3,3$
The denominators are different. Find a common denominator. LCD=$(x+3)(x-3)$ $\displaystyle \frac{x+3}{x-3}+\frac{x-3}{x+3}= \displaystyle \frac{x+3}{x-3}\times\frac{x+3}{x+3}+\frac{x-3}{x+3}\times\frac{(x-3)}{(x-3)}=$ $=\displaystyle \frac{(x+3)(x+3)+(x-3)(x-3)}{(x-3)(x+3)}\quad$ ... apply$: (a\pm b)^{2}=a^{2}\pm 2ab+b^{2}$ $=\displaystyle \frac{x^{2}+6x+9+x^{2}-6x+9}{(x-3)(x+3)}$ $=\displaystyle \frac{2x^{2}+18}{(x-3)(x+3)}$ $=\displaystyle \frac{2(x^{2}+9)}{(x-3)(x+3)},\quad x\neq-3,3$