Last night, I was working on an algorithm for calculating the square roots of numbers which worked by turning a fraction $\frac ab$ into $\frac{a+nb}{a+b}$, which converged to the square root of $n$. Then I thought about why it worked, and how I could make it work for higher-order roots.
I ended up getting to the algorithm of transforming $\frac ab$ into $\frac{a^{x-1} + n b^{x-1}}{a^{x-2} (a+b)}$ to make the number converge to the $x$'th root of $n$.
This should work because $\frac{a^{x-1} + n b^{x-1}}{a^{x-2} (a+b)}=\frac{y^{x-1}+n}{y^{x-1}+y^{x-2}}$ if $y=\frac ab$, and the difference between $y$ and $\frac{y^{x-1}+n}{y^{x-1}+y^{x-2}}$ gets smaller and smaller, making its limit $L=\frac{L^{x-1}+n}{L^{x-1}+L^{x-2}}$. When you simplify, this equation is actually $L^x=n$.
If you want to work on this, I would recommend working on the simplified version $\frac{y^{x-1}+n}{y^{x-1}+y^{x-2}}$.
I wrote a Mathematica code to check my equation for cube roots. The algorithm doesn't seem to work as it it supposed to.
Is my algorithm correct? When I went through it, it seems like it must work.
You can see the code of the function applied for cube roots at https://mathematica.stackexchange.com/questions/279846/ It only works for $\sqrt[3]{8}$.