解方程组（急求！！！！！！！！！！！！！）

三个方程等号两边同时加1
xy+x+y+1=2
yz+y+z+1=6
zx+z+x+1=3
所以(x+1)(y+1)=2
(y+1)(z+1)=6
(z+1)(x+1)=3
以上三式任意两个相乘再除以另一个有
(x+1)^2=1
(y+1)^2=4
(z+1)^2=9
就好做了

xy+x+y+1=2
yz+y+z+1=6
zx+z+x+1=3

(x+1)(y+1)=2........(1)
(y+1)(z+1)=6........(2)
(z+1)(x+1)=3........(3)

(1)*(2)*(3)
(x+1)^2*(y+1)^2*(z+1)^2=36
(x+1)(y+1)(z+1)=6.........(4)
(x+1)(y+1)(z+1)=-6........(5)

(4)/(1)..........z+1=3 z=2
(4)/(2)..........x+1=1 x=0
(4)/(3)..........y+1=2 y=1

(5)/(1)..........z+1=-3 z=-4
(5)/(2)..........x+1=-1 x=-2
(5)/(3)..........y+1=-2 y=-3

令X+Y+Z=A
XYZ=B
代入方程解