convergence rate of newton's method
Since you asked specifically about cubic, consider $f(x)=x+x^3$.
The Newton iteration would be $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=x_n-\frac{x_n+x_n^3}{1+3x_n^2}=\frac{2x_n^3}{1+3x_n^2}$$
It converges to the root $x=0$. So, for $1-3|x_n|^2&gt;1/2$,
$$\epsilon_{n+1}=|x_{n+1}-0|=\frac{2|x_n|^3}{|1+3x_n^2|}\leq \frac{2|x_n|^3}{1-3|x_n|^2}\leq 4|x_n-0|^3=4\epsilon_n^3$$