[tex]\frac{3}{(a-3)^2} +\frac{a+3}{9-3a} =\frac{3}{(a-3)^2} +\frac{a+3}{3*(3-a)}=\frac{3*3+(a-3)*(a+3)}{3*(a-3)^2} =\frac{9+a^2-9}{3(a-3)^2}=\frac{a^2}{3*(a-3)^2}[/tex]