Решите систему уравнений: x^3-y^3=8 x-y=2

X³-y³=(x-y)(x²+xy+y²)=8
x-y=2
$\left \{ {{2*( x^{2} +xy+ y^{2})=8 } \atop {x-y=2}} \right.$
$\left \{ {{( 2+y)^{2} +y*(2+y)+ y^{2} =4} \atop {x=2+y}} \right.$
$\left \{ {{4+2y+ y^{2}+2y+ y^{2} + y^{2}=4 } \atop {x=2+y}} \right. \left \{ {{3 y^{2} +4y=0} \atop {x=2+y}} \right. \left \{ {{y(3y+4)=0} \atop {x=2+y}} \right. \left \{ {{y=0 или y=- \frac{4}{3} } \atop {x=2+y}} \right.$
1) $\left \{ {{y=0} \atop {x=2}} \right.$
2) $\left \{ {{y=-1 \frac{1}{3} } \atop {x= \frac{2}{3} }} \right.$
Ответ: (0; 2); ( $-1 \frac{1}{3} ; \frac{1}{3}$ )