# 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.

• Do you want something that works even with irrational $n$? With regard to your question in general, if a factorization isn't obvious, or takes too long to help, you could always use something else to get the limit.
– J.G.
Feb 23, 2022 at 21:03
• As much as I'd like to use it, we aren't allowed any rules in my class so far, and we just have to use pure algebra, so I guess I'm looking for something that doesn't have to work when $n$ is irrational.
– user1029236
Feb 23, 2022 at 21:18

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 $$p_1/q_1.
Hence $$\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 (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 $$x\to a$$.
For the case : $$\lim_{x \to a} \dfrac{x^{\frac{3}{7}} - a^{\frac{3}{7}}}{x - a}$$ you can proceed by substitution : Let $$y = x^{\frac{1}{7}}$$ and $$b = a^{\frac{1}{7}}$$ then : $$\lim_{x \to a} \dfrac{x^{\frac{3}{7}} - a^{\frac{3}{7}}}{x - a} = \lim_{y \to b} \dfrac{y^3 - b^3}{y^7 - b^7}$$