I have been trying to solve the following difference equation for some time now : $$u^3(n+1) = a - b\cdot u^2(n) + u^3(n), \qquad a \ne 0 \ne b$$ I have tried various substitutions, simplifications but nothing did.

My last attempt was to transform it into a continuous differential equation by letting $u(n) = u(t_n)$, with $t_n = h + t_{n-1}$ and $t_0 = 0$ ie $t_n = h\cdot n$. From there I transformed the above equation in : $$\frac{u^3(t_{n+1}) - u^3(t_n)}{h} = \frac{a - b\cdot u^2(t_n)}{h}$$ $$\frac{u(t_{n+1}) - u(t_n)}{h} \cdot (u^2(t_{n+1}) + u(t_{n+1})\cdot u(t_n) + u^2(t_n)) = \frac{a - b\cdot u^2(t_n)}{h}$$ Taking the limit as $h\longrightarrow 0$, I get $$3u^2\cdot\frac{du}{dt_n} = \frac{a - b\cdot u^2}{h}$$

From here I can solve $\int\frac{3h\cdot u^2}{a-b\cdot u^2} du = \int dt_n$,

integrating gives me $$\frac {t_n} h = \frac{3\sqrt a \cdot tanh^{-1}(u\cdot \sqrt{\frac ba})}{b^{3/2}} - \frac{3u}{b} + C$$ but I'm interested in the inverse of this function, as in $u(t_n) = ...$

Does anybody know how to either solve anatically the initial difference equation or find an approximation?

Also is my continuous approach any good and is it possible to invert it? I tried using a polynomial expansion but it wouldn't converge quickly enough.