Finding iteratively approximation of cube root with square root operation It is given in the paper titled: A Simple Calculator Algorithm by: Lyle Cook, James McWilliam; the way to iteratively find using the cube root using the square root operation is to perform the set of two operations alternatively: (i) Take the square root twice; (ii) multiply by the original number, ...

I want to know first why this approach is deemed useful, i.e. on what basis.
It is also stated on the last page(#54) that a better way is to take the square root thrice, instead of twice. So, what is the reason for taking $y^\frac{1}{2\sqrt 2}$. (sorry, if not represented root properly, request correct way then)

Second, I tried to find the value obtained by program, by giving inputs : $57, 24, 8$. Request suggestions for improvements.
As explained in the paper, it is based on the infinite sum of geometric progression: $\sum_1^\infty(1/4)^n=1/3$. The number that you get after taking square root twice at each iteration is an approximation of the cubic root. I am not sure why they are suggesting to take the square root thrice, I do not think it's going to be an improvement.
An algorithm producing the cubic root of $x$ positive starts from $x_0=y_0=x$ and iterates the steps:


Compute $x_{k+1}=\sqrt{\sqrt{y_k}}$
Compute $y_{k+1}=x_{k+1}x$


Then, indeed, $$\lim x_k=x^{1/3}$$
because $$\frac14\left(1+\frac14+\frac1{4^2}+\frac1{4^3}+\cdots\right)=\frac13$$

Likewise, the algorithm starting from $u_0=v_0=x$ and iterating the step:


Compute $u_{k+1}=\sqrt{\sqrt{\sqrt{v_k}}}$
Compute $v_{k+1}=u_{k+1}x$


yields $$\lim u_k=x^{1/7}$$
because $$\frac18\left(1+\frac18+\frac1{8^2}+\frac1{8^3}+\cdots\right)=\frac17$$