Answer
$\frac{2(x^{2}+9)}{(x-3)(x+3)} ; x \ne 3,-3;$
Work Step by Step
$\frac{x+3}{x-3} + \frac{x-3}{x+3}$
$=\frac{(x+3)(x+3)+(x-3)(x-3)}{(x-3)(x+3)}; x \ne 3,-3;$
$=\frac{(x+3)^{2}+(x-3)^{2}}{(x-3)(x+3)}; x \ne 3,-3;$
$[(A+B)^{2} = A^{2}+2AB+B^{2}; (x+3)^{2} = x^{2}+6x+9;$
$ (A-B)^{2} = A^{2}-2AB+B^{2}; (x-3)^{2} = x^{2}-6x+9]$
$=\frac{x^{2}+6x+9+x^{2}-6x+9}{(x-3)(x+3)};x \ne 3,-3;$
$=\frac{2x^{2}+18}{(x-3)(x+3)}; x \ne 3,-3;$
$=\frac{2(x^{2}+9)}{(x-3)(x+3)} ;x \ne 3,-3;$
