Compare five ways of solving cubic equation by iterations (nested expressions) Say we have a depressed cubic equation in the general form:
$$x^3-bx-c=0$$
There are basically five ways of solving it by iterations. Let's consider them in no particular order (the names are my own).
1) The continued fraction / nested radical method:
$$x=\sqrt{b+\frac{c}{x}}=\sqrt{b+\cfrac{c}{\sqrt{b+\cfrac{c}{x}}}}=\sqrt{b+\cfrac{c}{\sqrt{b+\cfrac{c}{\sqrt{b+\cfrac{c}{x}}}}}}$$
Take some $x_0$ and solve the recurrence:
$$x_n=\sqrt{b+\frac{c}{x_{n-1}}}$$
2) The continued fraction / nested square method:
$$x=-\cfrac{c}{b-x^2}=-\cfrac{c}{b-\cfrac{c^2}{(b-x^2)^2}}=-\cfrac{c}{b-\cfrac{c^2}{\left( b-\cfrac{c^2}{(b-x^2)^2} \right)^2}}$$
3) The branching continued fraction method:
$$x=\cfrac{b}{x}+\cfrac{c}{x^2}=\cfrac{b}{\cfrac{b}{x}+\cfrac{c}{x^2}}+\cfrac{c}{\left( \cfrac{b}{x}+\cfrac{c}{x^2} \right)^2}$$
It can also be expressed in several different ways, like this one:
$$x=\cfrac{c+bx}{x^2}=\cfrac{c+b\cfrac{c+bx}{x^2}}{\cfrac{(c+bx)^2}{x^4}}$$
But anyway, for some $x_0$:
$$x_n=\cfrac{b}{x_{n-1}}+\cfrac{c}{x_{n-1}^2}$$
4) Nested cubic roots method:
$$x=\sqrt[3]{c+bx}=\sqrt[3]{c+b\sqrt[3]{c+bx}}=\sqrt[3]{c+b\sqrt[3]{c+b\sqrt[3]{c+bx}}}$$
5) And nested cubes method:
$$x=\frac{1}{b} \left(-c+x^3 \right)=\frac{1}{b} \left(-c+\frac{1}{b^3} \left(-c+x^3 \right)^3 \right)$$


They all work in different ways, so I wonder, what are general rules of using one method or the other? How to get all three roots by using them?

For example we have the equation with three real roots $(-3.65617,0.42297,3.23319)$:
$$x^3-12x+5=0$$
Let's take $x_0=1$. Then methods 1, 3 and 4 will converge to the third root, while methods 2 and 5 will converge to the second root.


Changing $x_0$ to other values will not lead to the first root, but some methods will not converge (mainly, method 3).

Let me try two general methods and have look if these show up in your list:
$$
f(x) = x^3 - bx - c = 0
$$
Newton-Raphson-Iteration
$$
x_{n+1} 
= x_n - \frac{f(x_n)}{f'(x_n)} 
= x_n - \frac{x_n^3 - b x_n - c}{3x_n^2 - b}
$$
Fixed point iteration:
$$
x^3 - bx - c = 0 \iff \\
x^3 - c = b x
$$
If $b \ne 0$ we can write this as
$$
x^3 - bx - c = 0 \iff \\
F(x) = \frac{x^3-c}{b} = x
$$
which leads to the iteration
$$
x_{n+1} = \frac{x_n^3-c}{b}
$$
The Netwon-Raphson I have not spotted, but the fixed point iteration is your number 5 (nested cubes). 
Newton-Raphson has quadratic convergence, so it is fast.
Fixed point iteration features attractive $\lvert F(x^*) \rvert < 1$ and not attractive fixed points $x^*$. In the latter case one could try to iterate $x = F^{-1}(x)$, because $(F^{-1})'(x) = 1 / F'(F^{-1}(x))$.
The properties of these iterations result from the properties of their general method.
Note that the different roots (up to three), need different start values for the iteration.
