If $(x^2+y^2+z^2)=2(x+z-1)$, then show that $x^3+y^3+z^3$ is constant and find its numeric value. I am trying to solve this question,

If $(x^2+y^2+z^2)=2(x+z-1)$, then show that $x^3+y^3+z^3$ is constant and find its numeric value.

I've tried this, 
$$x^2-2x + z^2-2z + 2 + y^2 = 0$$
$$ (x-1)^2 + (z-1)^2 + y^2 = 0$$
The left hand side can only become $0$ if $x=1$, $y=0$ and $z=1$, so the only solution is $x=z=1$ and $y=0$, which gives $x^3 + y^3 + z^3 = 2$.

  
*
  
*Have I solved it correctly?
  
*And is there any other method to solve it?
  


The OP wrote:  

The left hand side can only become $0$ if $x=1$, $y=0$ and $z=1$, so the only solution is $x=z=1$ and $y=0$, which gives $x^3 + y^3 + z^3 = 2$.  

I would add a justification that these values for $x$, $y$, and $z$ are unique. 
We know that each of the three terms, $(x-1)^2$, $(z-1)^2$ and $y^2$ are positive if we are dealing with Real Numbers. So, if any of them were non-zero, then at least one of the other terms would have to be negative for the equation to equal $0$. If we are dealing with Real Numbers, that is not possible. So each term must equal $0$. 
Once that justification is taken into account, including the assumption that ${x,y,z}\in \mathbb R$, then I think the OP has made a sound demonstration.
