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My level of perspective: I've been teaching myself mathematics mostly using Art of Problem Solving books, and various online resources. Have basically wrapped up the AoPS Prealgebra. Previously have worked through somewhat random concepts/techniques at varying levels/in varying branches of maths.

I was looking online for problems online and came across a problem that has an aspect that has been stumping me, a full explanation to follow.

$\frac{1}{(x^3+cx^{-3})^\frac{1}{3}}$, factoring $(x^{-3})^\frac{1}{3}$ = $\frac{x}{(x^6+c)^\frac{1}{3}}$

How that factoring results in the second expression with x in the numerator is thankfully clear to me.

My problem is this: The factoring would not be so unusual to me, if it weren't for the preservation of the rational exponent of the factor. What I am inclined to think is that the rule being applied is the power of a product rule. Then it makes sense, that by having the same rational exponent on the factored term as on its 'parent', I can apply the power of a product rule and multiplication of the factor by the terms from which it was initially factored can proceed as normal.

However, on a conceptual level, I'm having a real headache. My brain keeps trying to tell me that what 'should' happen is that the term that I factor out doesn't need that exponent, that is, I would just be factoring out $(x^{-3})$, instead of $(x^{-3})^\frac{1}{3}$. It seems strange to me that any terms 'inside' $(x^3+cx^{-3})^\frac{1}{3}$ , all carry a rational exponent property, so that when they are factored out, they carry that property with them. I can come up with all kinds of explanations for myself supporting that this is just an example of the product of powers rule, but I can't wrap my head around why any term that I should factor out of that sum would also have that rational exponent. I know this isn't a very clear problem, but perhaps someone sees/understands this agonizing caused by my intuition clashing with what (I think) I know.

Edit: I've been thinking that it makes sense that the factors be scaled by the exponent of the expression from which they are initially factored. In a manner of speaking, if one factored a term (here, reciprocal of x-cubed) and chose not to include the exponent as well (taking the cube-root of the factor), then the term would not be 'compressed' in the way that its parent was, and so it both could not have 'come from' that original parent (couldn't have been factored from), and, could not be multiplied by those terms directly, only after the terms had already had their cube-root taken.

I know someone is going to have to fix my Latex, I've never used it before.

Sincere thanks in advance for any advice.

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  • $\begingroup$ Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. $\endgroup$ – José Carlos Santos Jul 1 at 10:50
  • $\begingroup$ That - expression, factoring equation - needs to be explained. $\endgroup$ – William Elliot Jul 1 at 11:42
  • $\begingroup$ I think your paragraph beginning "Edit" hits the nail on the head, I mean, explains the matter perfectly. $\endgroup$ – Simon Jul 1 at 13:20
  • $\begingroup$ Do you have the same conceptual issue with what's below? If not, why? $$\large (x^9+cx^3)^2 = (x^3(x^6+c))^2 = x^6(x^6+c)^2$$ $\endgroup$ – Bill Dubuque Jul 1 at 14:33
  • $\begingroup$ That is a useful way to motivate the thinking here. I suppose it is that I could conceivably expand in the case of integer exponents, where I'm not sure that there is an analog with rational exponents (switching to using the radical sign is a substitution of notation that doesnt clarify the conceptual notion in quite the same way that expanding terms does, where one does get a sense that the exponential notation cleans up a lot of writing). $\endgroup$ – CrassTabloidHeadline Jul 2 at 14:44
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Step by step:

$$\frac{1}{(x^3+cx^{-3})^{1/3}}=\frac{1}{((x^{-3})(x^6+c))^{1/3}}=\frac{1}{(x^{-3})^{1/3}(x^6+c)^{1/3}}=\frac{1}{x^{-1}(x^6+c)^{1/3}} =\frac{x}{(x^6+c)^{1/3}}.$$

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    $\begingroup$ The power of products rule is clear. The difficulty I think is in the combination in this case of, the distributive property, the product of powers, and that the sum being factored has a rational exponent. None of this is unclear for me as a 'game of rules' but conceptually, it is hard to intuit why the factor also has the same rational exponent (explained without invoking product of powers rules is difficult). I think I understand if one considers that the factor must be of the same 'compressed-ness'(to the cube-root size) as the product from which it was factored. $\endgroup$ – CrassTabloidHeadline Jul 1 at 12:36
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    $\begingroup$ I'm very sorry I know this is all much too murky because I am sidestepping clear mathematics in favor of conceptual explanations, but for me, learning mathematics is the pleasure of this dissection, and I want to avoid taking anything at face value. I'm sure at some level of granularity the mathematical notation and rules will converge with the conceptual explanation my brain is longing for. $\endgroup$ – CrassTabloidHeadline Jul 1 at 12:37
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    $\begingroup$ I've tried to upovote your initial answer, Yves. Factoring inside the power expression at least notationally eases things for me (thanks Vedvart1 for highlighting that point). I understand the seeming absurdity of seeking wordy conceptual explanations for things which have perfectly concise, and correct, explanations using notation alone. $\endgroup$ – CrassTabloidHeadline Jul 1 at 12:59
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    $\begingroup$ @CrassTabloidHeadline I would recommend, if you get truly stuck on something, as weird as it feels just put it behind you and keep moving forward. At least in my experience, it helps to "blitzkrieg" the subject so to speak. From very simple math to more advanced or abstract stuff, I always try to understand as much as I can, but don't worry too much when I get stuck - because when I revisit the topic like a year later, I remember everything I learned the first time, and from everything I learned in the gap, all the holes just click, and I leave with a solid understanding. $\endgroup$ – Vedvart1 Jul 1 at 13:06
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    $\begingroup$ @Vedvart1 that is kind of you to say, and sage advice. I'll take that to heart. $\endgroup$ – CrassTabloidHeadline Jul 1 at 13:14

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