Срочно! Помогите с решение примера B= -

Решите задачу:

$\frac{64b^2 +128b +64}{b} : ( \frac{4}{b} +4)= \frac{64*(b^2 +2b +1) }{b} :( \frac{4+4b}{b} ) = \\ \\ = \frac{64*(b+1)^2}{b} : \frac{4(b+1)}{b} = \frac{64(b+1)^2}{b} * \frac{b}{4(b+1)} = \\ \\ = \frac{16(b+1)*1}{1*1} = 16(b+1) \\ \\ \\ b=- \frac{15}{16} \\ \\ 16*(- \frac{15}{16} +1) = 16 * \frac{1}{16} = \frac{16}{16} = 1$