I've asked a question some time ago about Computing square roots with arithmetic-harmonic mean but it turned out that this method is exactly the same as Newton's method (or Babylonian method) for square roots.
Now I figured out how to compute cube roots in the same way - and in the case of cube roots this method is apparently not the same as Newtons' method.
Let's introduce the following algorithm for some $x,y,z$:
$$a_0=\frac{x+y+z}{3},~~~~~~~b_0=\frac{xy+yz+zx}{x+y+z},~~~~~~~c_0=\frac{3xyz}{xy+yz+zx}$$
$$a_{n+1}=\frac{a_n+b_n+c_n}{3},~~~~~~~b_{n+1}=\frac{a_nb_n+b_nc_n+c_na_n}{a_n+b_n+c_n},~~~~~~~c_{n+1}=\frac{3a_nb_nc_n}{a_nb_n+b_nc_n+c_na_n}$$
We can obviously see that:
$$a_{n+1}b_{n+1}c_{n+1}=a_nb_nc_n=xyz$$
Thus, if the algorithm converges, we have:
$$\lim_{n \to \infty}a_n=\lim_{n \to \infty}b_n=\lim_{n \to \infty}c_n=\sqrt[3]{xyz}$$
For practical purposes it is much better to take $b_n$ as an estimate of the root, as can be shown by the following:
$$x=2,~~~~y=z=1$$
$$\begin{array}( n & a_n & b_n & c_n & \sqrt[3]{xyz} \\ 0 & 1.3333333333 & \color{blue}{1.25} & 1.2 & 1.2599210499 \\ 1 & 1.2611111111 & \color{blue}{1.2599}118943 & 1.2587412587 & 1.2599210499 \\ 2 & 1.2599214214 & \color{blue}{1.2599210499} & 1.2599206784 & 1.2599210499 \end{array}$$
This algorithm can be generalized for $n$th roots as well, but it will involve a lot more calculations at each step, since we will have to compute $n$ means from sums of products of $k$-tuples with $k=(1,2,\dots,n)$.
How do we prove this algorithm converges for any $x,y,z$?
Is it really different from Newton's method? If it is, could it be more efficient in some cases?