Помогите решить систему xy-x=2, xy^3-xy^2=8

$\left \{ {{xy-x=2} \atop {xy^3-xy^2=8}} \right. =\ \textgreater \ \left \{ {{xy-x=2} \atop {x(y^3-y^2)=8}} \right. =\ \textgreater \ \left \{ {{\frac{8y}{y^3-y^2}-\frac{8}{y^3-y^2}=2 \ (*)} \atop {x=\frac{8}{y^3-y^2}}} \right. \\ \\ (*) \ \ \frac{8y}{y^3-y^2}-\frac{8}{y^3-y^2}=2 \ | \ * (y^3-y^2) \ \ y \neq 0; 1.\\ \\ 8y-8=2y^3-2y^2 \ | \ : 2 \\ \\ y^3-y^2-4y+4 = 0 \\ \\ y^2*(y-1)-4*(y-1)=0 \\ \\ (y-1)*(y^2-4) = 0 \\ \\ (y-1)*(y-2)(y+2) = 0 =\ \textgreater \ y = б2.$
$y=2 \ : x=\frac{8}{2^3-2^2} \ = \frac{8}{4}=2 \\ \\ y=-2 \ : x=\frac{8}{(-2)^3-(-2)^2}=\frac{8}{-12} = -\frac{2}{3} \\ \\ \\ OTBET: (-\frac{2}{3}; -2), (2; 2).$ [