Solving the system $x^3+y= 3x+4$, $2y^3+z = 6y+6$, $3z^3+x=9z+8$ 
Solve the system $$\begin{equation} \label{equation1}
\begin{split}
x^3+y= 3x+4 \\
     2y^3+z = 6y+6 \\
3z^3+x=9z+8
\end{split}
\end{equation}$$

By the theorem of triviality, I assumed $x=y=z=k$ and then solved to fortunately get a solution. Since, it is a simultaneous system of equations the very solution I got is the only possible solution. But I can't understand how to do it by general method.
Any help is greatly appreciated.

Taking help from the previous answer, we can write the equations as
$$x^3-3x-2=(x+1)^2(x-2)=2-y$$
$$2(y^3-3y-2)=2(y+1)^2(y-2)=2-z$$
$$3(z^3-3y-2)=3(z+1)^2(z-2)=2-x$$
Multiplying the three, we get
$$6(x+1)^2(y+1)^2(z+1)^2(x-2)(y-2)(z-2) = -(x-2)(y-2)(z-2)$$
If none of $x,y,z$ is equal to 2 we can cancel $(x-2)(y-2)(z-2)$ on each side, which gives us $6(x+1)^2(y+1)^2(z+1)^2 = -1$, which is not possible. Hence, at least one of $x, y$ or $z$ equals 2. Putting this back to original equations, we get $(x,y,z) = (2,2,2)$.
