Требуется помощь с алгеброй, 99 баллов за решение и подробное объяснение. Желательно...

Решите задачу:

$( \frac{y}{4y+16}- \frac{y^2+16}{4y^2-64}- \frac{4}{y^2-4y} )* \frac{3y^2-24y+48}{y+4} =$

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

$=( \frac{y}{4(y+4)}- \frac{y^2+16}{4(y-4)(y+4)}- \frac{4}{y(y-4)} )* \frac{3(y-4)^2}{y+4}=$

$=(\frac{y(y-4)}{4(y+4)(y-4)}- \frac{y^2+16}{4(y-4)(y+4)}- \frac{4(y+4)}{y(y-4)(y+4)} )* \frac{3(y-4)^2}{y+4}=$

$=(\frac{y^2-4y}{4(y+4)(y-4)}- \frac{y^2+16}{4(y-4)(y+4)}- \frac{4y+16}{y(y-4)(y+4)} )* \frac{3(y-4)^2}{y+4}=$

$=\frac{y^2-4y-y^2-16-4y-16}{4(y+4)(y-4)}* \frac{3(y-4)^2}{y+4} =\frac{-8y-32}{4(y+4)(y-4)}* \frac{3(y-4)^2}{y+4}=$

$=\frac{-8(y+4)}{4(y+4)(y-4)}* \frac{3(y-4)^2}{y+4}=-\frac{2}{y-4}* \frac{3(y-4)^2}{y+4}=-\frac{6(y-4)^2}{(y-4)(y+4)}=-\frac{6(y-4)}{y+4}$