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!
Best Answer
Hint: $\;$ let $\,y_n = \dfrac{1}{x_n}\,$, invert both sides, then complete a cube.
$$y_{n+1}\color{red}{-1}=y_n^3-3y_n^2+3y_n\color{red}{-1}$$