Помогите пожалуйста решить

Объяснение:

$\left \{ {{x^3+2x^2y-xy^2-2y^3=0} \atop {x^2+y^2=8}} \right. \ \left \{ {{x*(x^2-y^2)+2y*(x^2-y^2)=0} \atop {x^2+y^2=8}} \right. \ \left \{ {{(x^2-y^2)*(x+2y)=0} \atop {x^2+y^2=8}} \right. \ \left \{ {(x-y)*(x+y)*(x+2y)=0}} \atop {x^2+y^2=8} \right.$ $1.\ x-y=0\ \ \ y=x\\x^2+x^2=8\\2x^2=8\ |:2\\x^2=4\\x_1=2\ \ \ x_2=-2\\y_1=2\ \ \ y_2=-2.$

$2.\ x+y=0\ \ \ y=-x\\x^2+(-x)^2=8\\x^2+x^2=8\\2x^2=8\ |:2\\x^2=4\\x_3=2\ \ \ x_4=-2\\y_3=-2\ \ \ y_4=2.$

$3.\ x+2y=0\ \ \ x=-2y\\(-2y)^2+y^2=8\\4y^2+y^2=8\\5y^2=8\ |:5\\y^2=1,6} \\y=б\sqrt{1,6 } \Rightarrow\\ y_5=\sqrt{1,6} \ \ \ y_6=-\sqrt{1,6} \\ x_5=-2*\sqrt{1,6} \ \ \ x_6=2*\sqrt{1,6} .$

Ответ: (2;2) (-2;-2) (2;-2) (-2;2) (-2√1,6;√1,6) (2√1,6;-√1,6).