Докажите тождество: (y^2+9/27-3y^2 + y/3y+9 - 3/y^2-3y): (3y+9)^2/3y^2-y^3= y/9y+27

$(\frac{y^2+9}{27-3y^2}+\frac{y}{3y+9}-\frac{3}{y^2-3y}):\frac{(3y+9)^2}{3y^2-y^3}$

$1) \frac{y^2+9}{27-3y^2}+\frac{y}{3y+9}-\frac{3}{y^2-3y}=\frac{y^2+9}{-3(y-3)(y+3)}+\frac{y}{3(y+3)}-\frac{3}{y(y-3)}= \\ \\ =\frac{y}{3(y+3)}-\frac{y^2+9}{3(y-3)(y+3)}-\frac{3}{y(y-3)}= \frac{y^2(y-3)-y(y^2+9)-9(y+3)}{3y(y-3)(y+3)} = \\ \\ =\frac{(y^3-3y^2)-(y^3+9y)-(9y+27)}{3y(y-3)(y+3)} =\frac{y^3-3y^2-y^3-9y-9y-27}{3y(y-3)(y+3)} = \\ \\ =\frac{-3y^2-18y-27}{3y(y-3)(y+3)} =\frac{-3(y+3)^2}{3y(y-3)(y+3)} =-\frac{y+3}{y(y-3)}$

$2) -\frac{y+3}{y(y-3)}:\frac{(3y+9)^2}{3y^2-y^3}=-\frac{y+3}{y(y-3)}:(-\frac{9(y+3)^2}{y^2(y-3)})= \\ \\ =\frac{y+3}{y(y-3)}*\frac{y^2(y-3)}{9(y+3)^2}=\frac{y}{9(y+3)}=\frac{y}{9y+27}$

ответ: тождество верно