Решите!Развёрнуто!!!

Решите задачу:

$( \frac{7}{x-3} -x-3)\cdot \frac{3-x}{x^2+8x+16} =( \frac{7-x(x-3)-3(x-3)}{x-3} )\cdot\frac{3-x}{(x+4)^2}= \\ \\ =( \frac{7-x^2+3x-3x+9}{x-3})\cdot\frac{3-x}{(x+4)^2}= \frac{-x^2+16}{x-3}\cdot\frac{3-x}{(x+4)^2}= \frac{-(x^2-16)}{x-3} \cdot\frac{3-x}{(x+4)^2}= \\ \\ = \frac{-(x+4)\cdot(x-4)\cdot(3-x)}{(x-3)\cdot(x+4)^2} = \frac{(x+4)\cdot(x-4)\cdot(x-3)}{(x-3)\cdot(x+4)^2} = \frac{x-4}{x+4}$