Help with convergence of sequence

Question

question is: by choosing right $x_0$ prove that sequence that written below is convergent to ${\sqrt a}$ $$x_{n+1} = \frac {x_n(x_n^2-3a)}{3x_n^2+a}$$ where $n\ge0$ and $a>0$.
I tried this way: $$x = \frac {x(x^2-3a)}{3x^2+a} \Rightarrow x(x^2-3a)=x(3x^2+a) \Rightarrow 2x(x^2+2a)=0 \, , \, since \, a>0 \quad x=0$$ and $x =0$ is only solution
Thanks in advance for helping me find out how to solve this

Answer 1

This (with the sign correction) is the Halley method applied to $f(x)=x^2-a$ or, in one way to construct that method, the Newton method to $f(x)=x^{3/2}-ax^{-1/2}$.

One can show relatively easily with the third degree binomial formula that $$ \frac{x_{n+1}-\sqrt a}{x_{n+1}+\sqrt a}=\left(\frac{x_{n}-\sqrt a}{x_{n}+\sqrt a}\right)^3 $$ from where the third order convergence follows.

See also Convergence behaviour of an iteration of a square root using Halley's method.