We have $0\le \sqrt{x^3-y}=z-1\implies z\ge 1\ $, $0\le \sqrt{y^3-z}=x-1\implies x\ge 1\ $ and $0\le \sqrt{z^3-x}=y-1\implies y\ge 1$ so $x,y,z\ge 1$ now we have $x^3-y=(z-1)^2=z^2-2z+1 \implies x^3-1=z^2-2z+y$, similarly $ y^3-1=x^2-2x+z$, and $ z^3-1=y^2-2y+x$ adding all these we have $(x^3+y^3+z^3)-3=(x^2+y^2+z^2)-(x+y+z)$, or $(x^3-x^2+x)+(y^3-y^2+y)+(z^3-z^2+z)=3$ Now we have $(x^3-x^2+x)=x(x(x-1)+1)\ge 1$, so $(x^3-x^2+x)+(y^3-y^2+y)+(z^3-z^2+z)\ge 3$ so as we have the equality, we have $x=y=z=1$