Find the general term of sequence Let $(x_{n})_{n\geq 1}$, $x_{n}\in \mathbb{R}$ be a sequence with $x_{1}\in [\frac{1}{2},\infty )$ and $$x_{n+1}=\frac{x_{n}^{3}}{3x_{n}^{2}-3x_{n}+1}$$
Find the general term of the sequence.
Attempts:
I though of solving the ecuation $3r^{3}-3r^{2}+r=r^{3}$ , that has the roots $r_{1}=0, r_{2}=\frac{1}{2}, r_{3}=1$
I aslo noticed that$$x_{n+1}=\frac{x_{n}^{3}}{(1-x_{n})^{3}+x_{n}^{3}}$$
What is the fomula for the general term? I can't find a conection with $x_{n+1}=\frac{x_{n}+\alpha }{x_{n}+\beta }$ or $x_{n+1}=\alpha x_{n}+\beta x_{n-1}$ or is  there another method?
Any help would be appreciated!
 A: Hint: $\;$ let $\,y_n = \dfrac{1}{x_n}\,$, invert both sides, then complete a cube.
$$y_{n+1}\color{red}{-1}=y_n^3-3y_n^2+3y_n\color{red}{-1}$$
A: I think the idea of identifying the cubes was a fantastic spot. Regarding solving the equation that you did : That's not the equation you need to solve, but a similar one. One should also be looking out for the well-definition of the RHS of the definition i.e. proving , for example, that $x_n$ is positive for all $n$. That is clear once one sees that $3x^2-3x+1 > 0$ for all $x >0$.
Once you do this, then you need to cleverly retain any created cubes : essentially , weed out the troublesome $3x_n^2-3x_n+1$. That isn't difficult to do. Indeed, if $$
x_{n+1} = \frac{x_n^3}{3x_n^2-3x_n+1}
$$
then subtracting $1$ from both sides$$
x_{n+1}-1 = \frac{x_n^3-3x_n^2+3x_n-1}{3x_n^2-3x_n+1} = \frac{(x_n-1)^3}{3x_n^2-3x_n+1}
$$
Now, IF $x_n \neq 1$ and $x_{n+1}\neq 1$, then dividing one equation by the other will retain the cubes as well as get rid of the troublesome $3x_n^2-3x_n+1$. When we do it, we get  $$\frac{x_{n+1}}{x_{n+1}-1} = \left(\frac{x_n}{x_n-1}\right)^3$$
which , provided we can retain the inequality to $1$ at each point, instantly leads to the following iterated situation :$$
\frac{x_{n+1}}{x_{n+1}-1} = \left(\frac{x_n}{x_n-1}\right)^3 = \left(\frac{x_{n-1}}{x_{n-1}-1}\right)^9 = \left(\frac{x_{n-2}}{x_{n-2}-1}\right)^{27} 
$$
and so on , till we get to :$$
\frac{x_{n+1}}{x_{n+1}-1} =  \left(\frac{x_1}{x_1-1}\right)^{3^{n}}
$$
which can be easily solved to yield a formula for $x_{n}$ only in terms of $x_1$ and $n$. Indeed, $$
x_{n+1} = \frac{\left(\frac{x_1}{x_1-1}\right)^{3^{n}}}{\left(\frac{x_1}{x_1-1}\right)^{3^{n}}-1}
$$

The IF case, requires solving an equation.
Indeed, IF $x_n =1$, then $x_{n+1} = 1$ by substitution, so the sequence is a constant $1$ afterwards.
On the other hand, if $x_{n+1} = 1$, then $x_n$ must satisfy $r^3 = 3r^2-3r+1$ so that $r = 1$ is forced from here. So $x_n = 1$ as well. In other words, if a $1$ ever appears in the sequence, then the only possibility is that the sequence is the constant sequence $1$.
Thus, we can complete our reasoning from here.
