Solve the system:

$$
\left\{\begin{array}{l}
y=x(4-x) \\
z=y(4-y) \\
x=z(4-z)
\end{array}\right.
$$
Making the substitution $x=2-2 u, y=2-2 v, z=2-2 w$, we obtain the system

$$
v=2 u^{2}-1
$$

$$
w=2 v^{2}-1, \quad, \quad
$$

$u=2 w^{2}-1$

According to problem $\underline{61282}$, the solutions of this system are $(1,1,1),(-1 / 2,-1 / 2,-1 / 2),\left(\cos 2 \pi / 9, \cos ^{4 \pi} / 9, \cos ^{8 \pi} / 9\right)$, ( $\cos 2 \pi / 7$, $\cos ^{4 \pi} / 7, \cos ^{8 \pi} / 7$ ) and all those obtained from them by cyclic permutations.

## Answer

$(0,0,0),(3,3,3),\left(4 \sin ^{2 \pi / 9}, 4 \sin ^{2} 2 \pi / 9,4 \sin ^{2} 4 \pi / 9\right),\left(4 \sin ^{2 \pi / 7}, 4 \sin ^{2} 2 \pi / 7,4 \sin ^{2} 3 \pi / 7\right)$ and all those obtained from them by cyclic permutations.