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.

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.
