Помогите срочно!!!!!!

Решите задачу:

$1) \frac{b^{3}}{b^{2}-8b+16} - \frac{b ^{2} }{b-4} = \frac{b^{3}}{(b-4)^{2}} - \frac{b ^{2} }{b-4} =\frac{b^{3}-b^{2}(b-4)}{(b-4)^{2}} =\frac{b^{3}-b^{3}+4b^{2}}{(b-4)^{2}} = \frac{4b^{2}}{(b-4)^{2}} \\ 2) \frac{b ^{2} }{b ^{2}-16 } - \frac{b}{b-4} =\frac{b ^{2} -b(b+4)}{b ^{2}-16 } =\frac{b ^{2} -b^{2}-4b}{b ^{2}-16 } =-\frac{4b}{b ^{2}-16 } \\ 3)\frac{4b^{2}}{(b-4)^{2}}:(-\frac{4b}{b ^{2}-16 })=-\frac{4b^{2}}{(b-4)^{2}}* \frac{b ^{2}-16 }{4b} = -\frac{b*(b+4)}{b-4} =\frac{b*(b+4)}{4-b}= \\$

$\frac{b^{2}+4b}{4-b}$