Решить систему уравнений:

Упростим первое уравнение

$x^3+y^3=7\\ (x+y)(x^2-xy+y^2)=7\\ (x+y)(x^2+2xy+y^2-3xy)=7\\ \\ (x+y)((x+y)^2-3xy)=7\\ \\ (x+y)^3-3xy(x+y)=7$

$\displaystyle \left \{ {{(x+y)^3-3xy(x+y)=7} \atop {xy(x+y)=-2}} \right. \to \left \{ {{(x+y)^3-3\cdot(-2)=7} \atop {xy(x+y)=-2}} \right. \to\\ \\ \\ \to \left \{ {{(x+y)^3=1} \atop {xy(x+y)=-2}} \right. \to \left \{ {{x+y=1} \atop {xy=-2}} \right. \to \left \{ {{x=1-y} \atop {y(1-y)=-2}} \right. \\ \\ y^2-y-2=0$
По т. Виета:
$y_1=-1;\,\,\,\,\,\,\, y_2=2$

$x_1=1-y_1=1+1=2\\ x_2=1-y_2=1-2=-1$

Ответ: $(2;-1),\,\,\,\, (-1;2).$