$(1) \Rightarrow x+y=z-1 $ $(1) \Rightarrow (x-y)(x+y)=(1-z)(1+z) \Rightarrow x-y=-1-z \Rightarrow x-y+z=-1 \Rightarrow x+y-z=x-y+z \Rightarrow y=z $ Problem becomes easy

magical wrote: $(1) \Rightarrow x+y=z-1 $ $(1) \Rightarrow (x-y)(x+y)=(1-z)(1+z) \Rightarrow x-y=-1-z \Rightarrow x-y+z=-1 \Rightarrow x+y-z=x-y+z \Rightarrow y=z $ Problem becomes easy that's only a case, you should consider the case when z-1=0 before dividing. so in that case z=1, then x=-y (from first equation) , then 2x^3=2(from third equation) so, x=1,y=-1,z=1 is also a solution. Otherwise, y=z (as you said) so x=-1 (1st equation) , and 2y^3=-2 (third equation) so y=-1=z, so the two solutions are : x=y=z=-1 and x=1,y=-1,z=1