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.
 A: Let $f(x)=-x^3+3x+2$, then the equations can be written as:
$$y-2=f(x), \quad z-2=2f(y), \quad x-2=3f(z)$$
This suggests simplification through the substitutions $a=x-2, b=y-2, c=z-2, g(t)=f(t+2)$ to get the equivalent
$$g(t)=-t(t+3)^2, \quad b=g(a), \; c=2g(b), \; a = 3g(c)$$
So we are trying to solve $a=3g(2g(g(a)))$, which is a polynomial with degree $27$ and has as many complex roots, which would in general need numerical methods to estimate.  We will look for only real roots in the following.
It is obvious $g(0)=0$ and hence if a root, say $a=0\implies b=c=0$ and this is always a solution (corresponds to $x=y=z=2$ for the original set of equations).
OTOH if $a$ is a non-zero real root, then noticing that $g$ always changes the sign of the argument, applying $g$ thrice (ignoring positive scaling factors) leads to an inconsistency. Hence there can be no non-zero real root.
A: 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 $(x,y,z) = (2,2,2)$.
