11. $$\frac{(3a-3c)^2}{9c^2 - 9a^2} = $$

$$\frac{(3a-3c)^2}{9c^2 - 9a^2} = \frac{(3(a-c))^2}{9(c^2-a^2)} = \frac{9(a-c)^2}{9(c-a)(c+a)} = \frac{(a-c)(a-c)}{(c-a)(c+a)} = \frac{-(c-a)(a-c)}{(c-a)(c+a)} = \frac{-(a-c)}{a+c} = \frac{c-a}{a+c}$$