Solve the system $x^3+y=3x+4$, $2y^3+z=6y+6$, $3z^3+x=9z+8$ 
Find the real solutions of the system below:
$$\begin{aligned} x^{3}+y &= 3x+4\\ 2y^{3}+z &= 6y+6\\ 3z^{3}+x &= 9z+8\end{aligned}$$

I wrote the system as:
$$x(x^{2}-3)=4-y$$
$$2y(y^{2}-3)=6-z$$
$$3z(z^{2}-3)=8-x$$
Then I tried to use the addition and multiplication of the equations above, but it didn't seem to be very useful.

We have
$$(y-2)=-(x-2)(x+1)^2\,,$$
$$(z-2)=-2(y-2)(y+1)^2\,,$$
and
$$(x-2)=-3(z-2)(z+1)^2\,.$$
Multiply the three equation above to get
$$(x-2)(y-2)(z-2)\big(1+6(x+1)^2(y+1)^2(z+1)^2\big)=0\,.$$
Since $1+6(x+1)^2(y+1)^2(z+1)^2\geq 1>0$, we conclude that
$$(x-2)(y-2)(z-2)=0\,,$$
whence $x=2$, $y=2$, or $z=2$.  However, this means $x=y=z=2$.
