1/2(1-y)+5/4y^2+4y+4=3y/y^3-1 решение

$\frac{1}{2(1-y)} + \frac{5}{4 y^{2}+4y+4} = \frac{3y}{ y^{3} -1} \\ \\ \frac{1}{2(1-y)} + \frac{5}{4( y^{2}+y+1)} = \frac{3y}{ (y -1)(y^{2} +y+1)} \\ \\ \frac{5}{4( y^{2}+y+1)} - \frac{1}{y-1} = \frac{3y}{ (y -1)(y^{2} +y+1)} \\ \\ \frac{5(y-1)-4(y^{2} +y+1)-4*3y}{(y -1)(y^{2} +y+1)} =0 \\ \\ \frac{5y-5-4y^{2}-4y-y-12y }{(y -1)(y^{2} +y+1)} =0 \\ \\ \frac{-4y^{2}-12y-5 }{(y -1)(y^{2} +y+1)} =0 \\ \\ y \neq 1$

-4y²-12y-5=0
4y²+12+5=0
D/4=6² - 4*5=16 (±4²)
y1=-6-4=-10
y2= -6+4= -2