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:


*

*$\sqrt{68}$

*$\sqrt[3]{401}$

*$\sqrt[7]{50}$

*$\sqrt[21]{675}$

*$\sqrt[n]{x}$

In a nutshell, the algorithm for square root works because of the identity:
$$ y =(a+x)^2 = a^2 + 2ax + x^2 = a^2 + (2a+x) x $$
In the algorithm you iteratively determine increasing (adding one decimal at the time) values for $a$ by looking for  the largest value of $x'$ at the appropriate decimal place so that $$y - a^2 \geq (2a+x') x'$$
You then essentially replace $a$ by $a+x'$ and continue for the next decimal place.
I suppose you could do this for e.g. cube roots by looking at
$$ y = (a+x)^3 = a^3 + (3 a^2+3x+x^2) x $$
but it looks rather nasty to carry out in practice.
