Hằng đẳng thức:\(a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(x^3+y^3+6xy=8\)
\(\Leftrightarrow\left(x^3+y^3+\left(-2\right)^3+6xy\right)=0\)
\(\Leftrightarrow\left(x+y-2\right)\left(x^2+y^2+4-xy+2y+2x\right)=0\)
\(\Leftrightarrow x+y=2\)