2021/033) Solve the system of equations

$xy = z - x - y\cdots(1)$

$yz = x - y - z\cdots(2)$

$zx = y - x - z\cdots(3)$

From (1) we have

Solution

x + y + xy = z

or xy + x + y + 1 = z + 1

or $(x+1)(y+1) = z+1\cdots(4)$

similarly from (2) and (3) get

 $(y+1)(z+1) = x+1\cdots(5)$

 $(z+1)(x+1) = y+1\cdots(6)$

 multiplying (4),(5),(56) we get

 $(x+1)^2(y+1)^2(z+1)^2 = (x+1)(y+1)(z+1)$

 so $(x+1)(y+1)(z+1) = 1$ or $(x+1)(y+1)(z+1) = 0$

  $(x+1)(y+1)(z+1) = 0$ gives x+ 1 = 0 or y + 1 = 0 or z + 1 = 0

  x+1 = 0 using (4) and (6)  y+ 1 = z+1 =0

 Similarly y + 1 = 0 gives  x+ 1 = z+1 =0

    and  z + 1 = 0 gives x+1 =  + 1 =0 

    so solution x = -1, y  = -1, z = -1

  product 1 along with (1), (2), (3) give x+1 = y+ 1 = z + 1 = 1 ( that is x= 1, y = 1, z = 1)

  or x + 1 = y+ 1 = -1 and z + 1 = 1( x=y=-2,z = 0)

 or y + 1 = z+ 1 = -1 and x + 1 = 1( y=z = -2,x = 0)

 or x + 1 = z+ 1 = -1 and y + 1 = 1( x=z=-2,y = 0)