Question

求数学大神来解答

Answer 1

0=x^3+y^3+3xy-1
=(x+y)^3-3x^2y-3xy^2+3xy-1
=(x+y)^3-3xy(x+y-1)-1
=[(x+y)-1][(x+y)^2+(x+y)+1]-3xy(x+y-1)
=(x+y-1)(x^2+2xy+y^2+x+y+1-3xy)
=(x+y-1)(x^2-xy+y^2+x+y+1)

所以x+y-1=0或者x^2-xy+y^2+x+y+1=0，

1. 若x+y-1=0 则为一条直线

2. 若x^2-xy+y^2+x+y+1=0，则y^2+(1-x)y+x^2+x+1=0,则Δ=(1-x)^2-4(x^2+x+1)=-3x^2-6x-3=-(x+1)^2<=0,有且仅当x=y=-1时成袜档立。运好肆

综上是一条直线x+y-1=0和一个点（-1，-1）旁轿

追问

谢谢