How to use the $x^n - a^n$ factoring pattern if $n$ is not an integer? $$x^n - a^n = (x-a)(x^{n-1} + x^{n-2}a \ + \ ... \ + xa^{n-2} + a^{n-1})$$
This works when $n$ is an integer, for example: $x^3 - a^3 = (x - a)(x^2 + xa + a^2)$.
This also works when factoring $x^1 - a^1$, for example: $x-a = (x^\frac13 - a^\frac13)(x^\frac23 + x^\frac13a^\frac13 + a^\frac23)$.
My question is if you had that first factor, how would you go about finding that second factor when it's a weird fraction? i.e something like $x-a = (x^\frac37 - a^\frac37)(...)$, how would you find that (...) part?
The reason I'm asking this is for questions where I have to differentiate weird equations with first principles, like $\frac{d}{dx}x^\frac37$. In class, when we got $\lim \limits_{x\to a} \frac{x^\frac37 - a^\frac37}{x-a}$, we were told to factor the numerator to get something like $(x^\frac17 - a^\frac17)(...)$ and in addition factor the denominator to get $(x^\frac17 - a^\frac17)(...)$ so that you could remove that term, etc. Is there any term I could multiply $\lim \limits_{x\to a} \frac{x^\frac37 - a^\frac37}{x-a}$ with to get $\lim \limits_{x\to a} \frac{x-a}{(x-a)(...)}$ instantly? What's the strategy for finding it?
Any help is appreciated, thanks.

If $n=p/q$ is a (positive) rational, you already know how to do:
$$
x^n-a^n=(x^{1/q})^p-(a^{1/q})^p= (x^{1/q}-a^{1/q})\cdot \sum_{i=0}^{p-1}(x^{1/q})^i(a^{1/q})^{p-1-i}.
$$
Now suppose that $n$ is a (positive) irrational.
Note that, for each $x\neq a$, the function $x^y-a^y$ has a derivative (in $y$) equals to $\log x\cdot  x^y-\log a\cdot a^y$ hence has a constant (nonzero) sign in a neighborhood $U$ of $n$.
Approximate $n$ with rationals: find rationals $p_1/q_1,p_2/q_2 \in U$ such that $p_1/q_1<n<p_2/q_2$.
Hence
$$
\frac{x^{p_1/q_1}-a^{p_1/q_1}}{x-a} \le \frac{x^n-a^n}{x-a}\le \frac{x^{p_2/q_2}-a^{p_2/q_2}}{x-a}
$$
for $x>a$ and viceversa for $x<a$ (or the opposite). But you know how to factorize the first term and the last term. Therefore you can find the limit of the middle one as $x\to a$.
