Is there a process, similar to long division, to do nth roots where n is any positive integer? I did not even remember how to do square roots from high school, but I vaguely recall it was similar to long division. (Thankfully I remember how to do long division.) I just went to youtube and refreshed my memory on how to do square roots.
Is there a "long root" process to do nth roots? N can be any integer of 2 or more. (N could be 1 too but that is trivial.) The radicand can be any positive number. The radicand does not hafta be a perfect square, perfect cube, or perfect n-power of anything.
Btw, I am aware of factorization. So $\sqrt{153} = \sqrt{9 * 17} = 3\sqrt{17}$. In this case, what I want to do is something like 2 into 17, but instead of long division, use a "long root" process for putting 2 into 17. The process would go on forever, much like 25/7 goes on forever because the remainder never "settles". I would just stop when I get, say, 3 decimal places or however many I think is accurate enough.
Examples:


*

*$\sqrt{68}$

*$\sqrt[3]{401}$

*$\sqrt[7]{50}$

*$\sqrt[21]{675}$

*$\sqrt[n]{x}$

I'm having trouble understanding exactly what you are asking about but I'll focus on the following "The process would go on forever, much like 25/7 goes on forever because the remainder never "settles". I would just stop when I get, say, 3 decimal places or however many I think is accurate enough."

You might want to grab a book on introductory real analysis. R. P. Burn's Numbers and Functions has very good chapters on sequences and completeness that will satisfy your needs.
For example, for any $a$, $0\leq a-\frac{\lfloor a10^n \rfloor}{10^n}<\frac{1}{10^n}$ ($\lfloor x \rfloor$ here is called floor $x$, which is the greatest integer less than $x$). From this we can deduce that for any number $a$ there is a sequence of rational numbers which tends to it. This is because the sequence $(1/10^n)$ tends to zero and so by the squeeze rule for null sequences, the sequence $(\frac{\lfloor a10^n \rfloor}{10^n}-a)$ tends to zero, so by the definition of limit we see how the result is proved. Remember that $\frac{\lfloor a10^n \rfloor}{10^n}$ is always rational.
The chapter on completeness will go over how every nth root of a number exists and is unique.
Hope that points you in the right direction.
