\noindent{\bf Answer}:

$[\textrm{Cubic}]$ 

One solution is given by (there may be others)
$$x_{n+1} = g(x_n), \quad g(x) = x - \frac{f(x)}{f'(x)} 
- \frac{f''(x)}{2f'(x)}\left[\frac{f(x)}{f'(x)}\right]^2. \quad \text{{\bf (5 pts)}}$$
% \underline{Part 2} (8 pts): 
Check that $g'(p)=g''(p)=0$ and $g^{(3)}(p) \neq 0$, 
and then compute 
$$\lim_{n \rightarrow \infty} \frac{|p_{n+1}-p|}{|p_n-p|^3} 
= \frac{|g^{(3)}(p)|}{3!}.  \quad \text{{\bf (10 pts)}}$$

$[\textrm{Quadratic}]$ 

$$x_{n+1} = g(x_n), \quad g(x) = x - \frac{f(x)}{f'(x)}. \quad \text{{\bf (2 pts)}}$$
%\underline{Part 2} (4 pts): 
Check that $g'(p)=0$ and $g''(p) \neq 0$, and then compute 
$$\lim_{n \rightarrow \infty} \frac{|p_{n+1}-p|}{|p_n-p|^2} 
= \frac{|g''(p)|}{2!}. \quad \text{{\bf (5 pts)}} $$