Question

已知x^3+y^3+3xy=1,求x+y

已知x^3+y^3+3xy=1,求x+y=—— 急求！

Answer 1

x³+y³+3xy=1
(x+y)³-3x²y-3xy²+3xy-1=0
[(x+y)³-1]-3xy(x+y-1)=0
(x+y-1)[(x+y)²+(x+y)+1]-3xy(x+y-1)=0
(x+y-1)[(x+y)²+(x+y)+1-3xy]=0
(x+y-1)(x²-xy+y²+x+y+1)=0
(x+y-1)[(1/2)(x²+2x+1)+(1/2)(y²+2y+1)+(1/2)(x²-2xy+y²)]=0
(x+y-1)[(x+1)²+(y+1)²+(x-y)²]=0
x+y-1=0或(x+1)²+(y+1)²+(x-y)²=0
x+y-1=0 x+y=1
(x+1)²+(y+1)²+(x-y)²=0，平方项恒非负，三非负项之和=0，三非负项均=0
解得x=-1 y=-1
x+y=(-1)+(-1)=-2
综上，得x+y=1或x+y=-2