How to get the cubic iteration of the Newton method? I have the taylor expansion: 

$$f(x+h)=f(x)+hf'(x)+ \frac 1 2 h^{2}f''(x)+O(h^3) $$

and I am trying to get the cubic iteration of the the Taylor method: 
$ x_{n+1}=x_{n}-f'(x_{n})(1-\frac{\sqrt{(1-2f"(x_{n})f(x_{n})})}{f"(x_{n})} )$
I have tried setting $f(x+h)=0$ and solving for h but it doesn't seem to be working 
 A: Improving on Donald Splutterwit, you want the smaller of the two solutions, so
\begin{align}
h&=-\frac{f'(x)}{f''(x)}\left(1-\sqrt{1-2\frac{f(x)f''(x)}{f'(x)^2}}\right)\\
&=-\frac{f'(x)}{f''(x)}\frac{2\frac{f(x)f''(x)}{f'(x)^2}}{1+\sqrt{1-2\frac{f(x)f''(x)}{f'(x)^2}}}\\
&=-\frac{2f(x)}{f'(x)}\left(1+\sqrt{1-2\frac{f(x)f''(x)}{f'(x)^2}}\right)^{-1}
\end{align}
which is Halley's original method. Only later was the square root replaced by the linear Taylor polynomial to get the "modern" Halley method that can also be written as
$$
x_+=x+\frac{d(1/f)^{(d-1)}(x)}{(1/f)^{(d)}(x)}
$$
for $d=2$ ($d=1$ gives the usual Newton method).

Another method to get a third order $h$ is to multiply
$$
0=f(x)+(f'(x)+\tfrac12f''(x)h)h+O(h^3)
$$
with $(f'(x)-\tfrac12f''(x)h)$ to get
$$
0=f(x)f'(x)+\Bigl(f'(x)^2-\tfrac12f(x)f''(x)\Bigr)h+O(h^3)
$$
which now is linear in $h$ and can be solved to
$$
h=-\frac{f(x)f'(x)}{f'(x)^2-\tfrac12f(x)f''(x)}
$$
which is again the increment of the Halley method.
A: Let $f(x+h)=0$, so we have $h^2f''(x)+2hf'(x)+2f(x)=0$. Now solve this as a quadratic in $h$
\begin{eqnarray*}
h= \frac{-f'(x) \pm \sqrt{ (f'(x))^2 -2 f(x)f''(x)}}{f''(x)} 
\end{eqnarray*}
Now let $h=x_{n+1}-x_n$ and $x=x_n$ and we have the desired result.
