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

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

Ответ: $$\frac{c-a}{c+a}$$