Find all the solutions of this equation:$x+3y=4y^3 \ , y+3z=4z^3\ , z+3x=4x^3 $ in reals. Find all the solutions of this equation
$x+3y=4y^3 \ , y+3z=4z^3\ , z+3x=4x^3 $
My attempt:In the hint of the question it was written that show $x,y,z \in [-1,1]$ then add all equations and conclude $(x,y,z)=(-1,-1,-1),(0,0,0),(1,1,1)$ are the solutions.I can show that $x,y,z \in [-1,1]$like below:
By symmetry consider $x \ge y \ge z$ then if $x>1$ we have:
$4x^3-4x>0 \Rightarrow 4x^3-3x>x \Rightarrow z>x$
which is wrong.Now consider $x<-1$ we have:
$x<-1 \Rightarrow y<-1 \Rightarrow 4y^3-4y<0 \Rightarrow 4y^3-3y<y \Rightarrow x<y$
By the same way we can get $x,y,z \in [-1,1]$.By adding the equations we can get:
$x+y+z=x^3+y^3+z^3$
But what should I do now?
 A: Preliminary lemma: knowing the classical relationship 
$\tag{0}\cos(3t)=4\cos^3(t)-3\cos(t)$ (recalled by @Achille Hui)
and knowing the property/definition of Chebyshev polynomials:
$T_n(cos(t))=cos(nt)$, we  have :
$$T_{27}(x)=T_3(T_3(T_3(x))).$$
Proof : $cos(27t)=cos(3(3(3t))).$

The system of 3 equations can be written under the form 
$$x=f(y), y=f(z), z=f(x)$$
with $f(t):=t^3-3t$.
Thus, using the lemma: $f(f(f(t))=T_{27}(t)$.
It is clear that $x,y,z$ are solutions of  fixed point relationships :
$$u=T_{27}(u)$$
knowing that the curve of any $T_n$ maps $[0,1]$ onto [0,1], and in this case, that $T_{27}(1)=1.$ 
Thus, as shown on the graphical representation below, the abscissas of intersection points with the straight line with equation $y=x$ are confined in $[-1,1]$. 

Remark: there are 27 intersection points. Thus we await 27 solutions, knowing that, once we have chosen $x$, there is no more choice for $y=f(x)$ and for $z=f(f(x))$ among these roots.
Let us find an explicit expression for these systems $(x,y,z)$ of roots, under a trigonometric form.
These roots, being in $[-1,1]$, can be represented resp. as 
$$x=\cos(a), y=\cos(b), z=\cos(c)$$
for certain values of $a,b,c$. Using (0), we can rewrite the initial system into the equivalent form:
$$\cos(a)=\cos(3b), \ \ \cos(b)=\cos(3c), \ \ \cos(c)=\cos(3a)$$
which is equivalent mod $2 \pi$ to:
$$\tag{1} a=s_1 \ 3b, \ \ b=s_2 \ 3c, \ \ c=s_3 \ 3a$$
(where $s_k=\pm1$). Taking the product of these 3 congruences, the only possible solutions are 
$$abc=27abc \ \text{mod} \  2\pi \ \ \text{or}  \ \ abc=-27abc  \ \text{mod} \  2\pi$$
giving $26 abc=k 2 \pi$ or $28 abc=k 2 \pi$ for a certain integer $k$, i.e., $abc=\dfrac{k 2\pi}{26}$ or $abc=\dfrac{k 2 \pi}{28}.$
Now, due to (1), $abc=s 27 a$ (where $s=\pm1$); whence two categories of cases :
$$\tag{2}\begin{cases}a&=&s\dfrac{27}{26} k 2 \pi=s\dfrac{1}{26} k 2 \pi \\
a&=&s\dfrac{27}{28} k 2 \pi=s \dfrac{1}{28} k 2 \pi\end{cases}$$
(where $s=\pm1$). It means that taking in (2), successively, $k=0,1, \cdots $, we will obtain the different solutions for $a$. Due to (2), these results for $a$ will generate all the solutions under the form: 
$$\tag{3}(x,y,z)=(\cos(a),\cos(9a),\cos(3a)) \ \text{with either} \ a=\dfrac{k \pi}{13} \ or \ a=\dfrac{k \pi}{14},$$  
with $k \in \mathbb{Z}$. (note that the signs have been dropped because $\cos(-t)=\cos(t)$).
In conclusion, one can verify that relationships (3) do not generate $26+28$ solutions, but in fact $27$, due to a certain number of "coincidences". 
Among the solutions, one finds $(x,y,z) = (0,0,0), (1,1,1)$ and $(-1,-1,-1)$.
A: Brute force. If we use $y=4z^3-3z$ and $z=4x^3-3x$ in $x=4y^3-3y$, we obtain a huge polynomial of $27$th degree in $x$ which can be factorized (by some not-human algebraic manipulator) as
$$4x(x-1)(x+1)(8x^3+4x^2-4x-1)(8x^3-4x^2-4x+1)\\(64x^6-112x^4+56x^2-7)(64x^6-32x^5-80x^4+32x^3+24x^2-6x-1)\\(64x^6+32x^5-80x^4-32x^3+24x^2+6x-1).$$
If we are interested in the integers solutions then we have only $x=0$, $x=1$ and $x=-1$ because the other factors do not have integer roots (check each constant term). 
Now $x\in\{-1,0,1\}$ implies that $z=4x^3-3x=x$, $y=4z^3-3z=x$ and therefore the integer solutions of the system are: $(0,0,0)$, $(1,1,1)$ and $(-1,-1,-1)$.
A: I am assuming that you want integer solutions.  First note that if any of $x$, $y$, and $z$ is zero, then all of them are.  Suppose now that they are all nonzero.  Then, you have relations $y\mid x$, $z\mid y$, and $x\mid z$, whence $|x|=|y|=|z|=:t$.  Then, $x+3y=4y^3$ implies that $$4t\geq |x+3y|=\left|4y^3\right|=4t^3\,,$$
making $t=1$ the only possibility.  The rest is straightforward, and the only solutions are $(x,y,z)=(\alpha,\alpha,\alpha)$ with $\alpha\in\{-1,0,+1\}$.
