x^3+y^3+z^3-3xyz
=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)
=3[(x+y+z)^2-3(xy+yz+zx)]
3-3xyz=3[(x+y+z)^2-3(xy+yz+zx)]
=>1-xyz=9-3(xyyz+zx)
=>xyz-3(xy+yz+zx)+8=0
=>xyz-3(xy+yz+zx)+9(x+y+z)-27=-8
=>(x-3)(y-3)(z-3)=-8
(x,y,z)=(-5,4,4),(4,-5,4),(4,4,-5),(1,1,1)
(有別解法在提供吧)