lpetrich
Contributor
So one gets possible x values -1/3, 1/2, 1, and 3. By symmetry, the other variables may also have these values.
Substituting in x and repeating this procedure to find y, I find
x = 1 -> y = 1
x = 1/2 -> y = 1/2
x = 3 -> y = 3 or y = -1/3
x = -1/3 -> y = 3
In each case, there is one value of z that completes the solution. (x,y,z) is
(1,1,1)
(1/2, 1/2, 1/2)
(3, 3, -1/3)
(3, -1/3, 3)
(-1/3, 3, 3)
Substituting in x and repeating this procedure to find y, I find
x = 1 -> y = 1
x = 1/2 -> y = 1/2
x = 3 -> y = 3 or y = -1/3
x = -1/3 -> y = 3
In each case, there is one value of z that completes the solution. (x,y,z) is
(1,1,1)
(1/2, 1/2, 1/2)
(3, 3, -1/3)
(3, -1/3, 3)
(-1/3, 3, 3)