Система уравнений с двумя неизвестными |x^2*y-x*y^2=6 |x^3-y^3=-9

$\left \{ {{x^2y-xy^2=6} \atop {x^3-y^3=-9}} \right.$

Выносим общий множитель
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$\left \{ {{xy(x-y)} \atop {(x-y)(x^2+xy+y^2)=-9}} \right. \to \left \{ {{-3xy(x-y)-2(x^2+xy+y^2)(x-y)=-3\cdot 6-2(-9)} \atop {(x^2+xy+y^2)(x-y)=-9}} \right.$

$\left \{ {{3xy+2(x^2+xy+y^2)=0} \atop {x^3-y^3=-9}} \right. \to \left \{ {{2x^2+5xy+2y^2=0} \atop {x^3-y^3=-9}} \right. \to \left \{ {{2x^2+xy+4xy+2y^2=0} \atop {x^3-y^3=-9}}\right. \to \\ \to \left \{ {{(2x+y)(x+2y)=0} \atop {x^3-y^3=-9}} \right.$
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Имеем 2 системы

$\left \{ {{2x+y=0} \atop {x^3-y^3=-9}} \right.$ и $\left \{ {{x+2y=0} \atop {x^3-y^3=-9}} \right.$

Случай 1.
$\left \{ {{2x+y=0} \atop {x^3-y^3=-9}} \right. \to \left \{ {{y=-2x} \atop {x^3-(-2x)^3=-9}} \right. \\ x^3+8x^3=-9 \\ 9x^3=-9 \\ x^3=-1 \\ x_1=-1 \\ y_1=-2\cdot (-1)=2$

случай 2.
$\left \{ {{x+2y=0} \atop {x^3-y^3=-9}} \right. \to \left \{ {{x=-2y} \atop {(-2y)^3-y^3=-9}} \right. \\ -8y^3-y^3=-9 \\ -9y^3=-9 \\ y_2=1 \\ x_2=-2$

ответ: $(-1;2),\,\,\,\,(-2;1).$