Find the possible value from the following. Find the possible value from the following.
I'm not able to end up on a concrete note, as I'm unable to get the essence of question, still not clear to me. 

$x$, $y$, $z$ are distinct reals such that $y=x(4-x)$, $z=y(4-y)$, $x=z(4-z)$. The possible values of $x+y+z$ is:
$$\begin{array}{l} A.\ 4 && C.\ 7 \\ B.\ 6 && D.\ 9 \end{array}$$

 A: Expand $y$ and $z$ to get a polynomial in $x$ by itself.
$$
x = y(4-y)(4-y(4-y))=x(4-x)(x-2)^2(4-x(4-x)(x-2)^2)
$$
In fact it's easier if you write $a=x-2,b=y-2,c=z-2$ then the original expressions become $b=2-a^2$, etc. and you get
$$
a = 2-(2-(2-a^2)^2)^2
$$
This gives a degree 8 polynomial in $a$. Two of the roots are integers you can find by inspection. Factoring them out leaves a degree 6 polynomial which factors as two cubics. Then if the roots of each cubic give a solution for $(a,b,c)$ (they do), then you can read off their sum from the coefficients. As I said in the comments, though, this seems to give two different solutions from the choices.
A: Composing the functions, we get
$$
\begin{align}
0
&=x^8-16x^7+104x^6-352x^5+660x^4-672x^3+336x^2-63x\\
&=x(x-3)(x^3-7x^2+14x-7)(x^3-6x^2+9x-3)\tag{1}
\end{align}
$$
The roots $x=0$ and $x=3$ lead to indistinct $x$, $y$, and $z$.
$x^3-7x^2+14x-7$ has 3 real roots in $[0,4]$ whose sum is $7$ (the negative of the coefficient of $x^2$).
$x^3-6x^2+9x-3$ has 3 real roots in $[0,4]$ whose sum is $6$ (the negative of the coefficient of $x^2$).
$x$, $y$, and $z$ all satisfy $(1)$.
$t(4-t)$ rotates the roots of the cubics.
Thus, the possible values of $x+y+z$ are $6$ and $7$.
A: Hint: If you graph $y=f(x)$ versus $x$, which is the same function so that $z=f(y)$ and $x=f(z)$, you will see that it is a parabola opening downwards with maximum at $(2,4)$ and $x$-intercepts at $0$ and $4$. Then try some integer values between the intercepts and you should quickly get the answer.
A: Alternative hint: look for a fixed point. 
