do x+y-2=0 ⇒\(\left\{{}\begin{matrix}x+y=2\\y-2=-x\end{matrix}\right.\)
ta có A = x3 + x2y - 2x2 - xy - y2 + 3y + x + 2016
= x3 + (x2y - 2x2) - (xy + y2) + 3y + x + 2016
= x3 + x2(y - 2) - y(x + y) + 3y + x + 2016
= x3 + x2.(-x) - 2y + 3y + x +2016
= x3 + (-x)3 + ( -2y + 3y) + x + 2016
= 0 + y + x + 2016
= (x + y) + 2016
= 2 + 2016
= 2018