Find all the value of $(x, y, z)$ with $x, y, z \in \Bbb R$ $$x^2 + 4 = y^3 + 4x - z^3$$
$$y^2 + 4 = z^3 + 4y - x^3$$
$$z^2 + 4 = x^3 + 4z - y^3$$
What's the proper method to solve this question? I've tried any method but still not get the answer.

$$x^2 + 4 = y^3 + 4x - z^3 \implies x^2 -4x +4 = y^3-z^3 \implies (x-2)^2 = y^3-z^3 \quad\text{ (1.)}$$
$$y^2 + 4 = z^3 + 4y - x^3 \implies y^2 -4y +4 = z^3 -x^3 \implies (y-2)^2 = z^3-z^3 \quad\text{ (2.)}$$
$$z^2 + 4 = x^3 + 4z - y^3 \implies z^2 -4z +4 = x^3-y^3 \implies (z-2)^2 = x^3-y^3 \quad\text{ (3.)}$$
Adding all three equations:
$$(x-2)^2 + (y-2)^2 + (z-2)^2 = 0$$
$$x-2 = 0 \,\,,\,\, y-2 = 0\,\,,\,z-2=0$$
$$\boxed{x=y=z=2}$$
