<s>[INST] prove that $x_{n+1}=\frac{x_n(x_n^2+15)}{3x_n^2+5}$ is cubic order of convergence near $x_0=\sqrt{5}$ To solve the equation
$$x^2-5=0$$
There exitsts a iteration method
  $$x_{n+1}=\frac{x_n(x_n^2+15)}{3x_n^2+5}$$ 
I know that it is cubic convergence but I don't know how to prove it.
I have tried the following
$$
\begin{align}
e_{k+1}&=x_{n+1}-\sqrt{5} \\ &=\frac{x_n(x_n^2+15)}{3x_n^2+5}-\sqrt{5} \\ &=\frac{\frac{1}{3}x_n(3x_n^2+5+40)}{3x_n^2+5}-\sqrt{5} \\ &=\frac{1}{3}x_n(1+\frac{40}{3x_n^2+5})-\sqrt{5} \\ &=\frac{\frac{40}{3}x_n}{3x_n^2+5}+\frac{1}{3}x_n-\sqrt{5}
\end{align}
$$
I'm trying to extract $e_n=x_n-\sqrt{5}$ from $e_{n+1}$ in an effort of proving $lim \frac{e_{n+1}}{e_n^3}\rightarrow C$ . But it seems that I'm not on the right way. 
Do you have any idea? Any discussion is appreciated. Thanks in advance.
 [/INST] In general, Halley's method for the square root reads as
$$
x_{n+1}=x_n·\frac{x_n^2+3a}{3x_n^2+a}
$$
The difference to the square root develops then as
$$
x_{n+1}-\sqrt{a}=\frac{x_n^3+3x_n\sqrt a^2 - 3x_n^2\sqrt a-\sqrt a^3}{3x_n^2+a}
=\frac{(x_n-\sqrt a)^3}{3x_n^2+a}
$$
which gives you the third order convergence.
</s>