| x+y |
| z |
| x+z |
| y |
| y+z |
| x |
| x+y |
| z |
| x+z |
| y |
| y+z |
| x |
∴x+y=kz，x+z=ky，y+z=kx，
∴k（x+y+z）=2（x+y+z），
当x+y+z=0时，则x+y=-z，x+z=-y，y+z=-x，
原式=
| −z•(−x)•(−y) |
| xyz |
=-1；
当x+y+z≠0时，则k=2，
原式=
| kz．kx．ky |
| xyz |
=k3=8．
故答案为：-1或8．
| x+y |
| z |
| x+z |
| y |
| y+z |
| x |
| (x+y)(y+z)(z+x) |
| xyz |
| x+y |
| z |
| x+z |
| y |
| y+z |
| x |
| x+y |
| z |
| x+z |
| y |
| y+z |
| x |
| −z•(−x)•(−y) |
| xyz |
| kz．kx．ky |
| xyz |