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Can someone factor this for me? $(x^{\frac{n}3}-a^{\frac{n}3})$

I am stuck on it. Let n be any natural number.

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closed as unclear what you're asking by Mark Bennet, user2345215, user86418, mookid, vonbrand Apr 3 at 2:26

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

    
It depends a bit on what $n$ is. If $n=3$ you get $x-a$. If $n=6$ you get $(x+a)(x-a)$. If $n$ is not a multiple of $3$, this is not a polynomial (per title, but not tag). –  Mark Bennet Apr 2 at 22:44
    
cause I have that, over (x-a) and I am trying to reduce it and I can't seem to get anywhere. –  Jack Apr 2 at 22:48
    
Because that's about as simple as it gets for arbitrary $n$. –  Graham Kemp Apr 2 at 23:11
    
Maybe you can say what the larger problem is. Are you taking the limit of that fraction as $x\rightarrow a$, for example? –  MPW Apr 2 at 23:21
1  
Note that $$\forall n\in \mathbb N\forall \alpha,\beta\in \mathbb R\left(\alpha^n-\beta^n=(\alpha-\beta)\left(\alpha ^{n-1}+\alpha^{n-2}\beta+\ldots +\alpha\beta^{n-2}+\beta^{n-1}\right)\right).$$ Take $\alpha=x^{1/3}$ and $\beta=a^{1/3}$. –  Git Gud Apr 2 at 23:29

1 Answer 1

up vote 2 down vote accepted

From your comment, it looks like you are being asked to find a "difference quotient". Subject the condition that the numbers being raised to fractional exponents need to be positive, you can treat this as

$$ \frac{x^{n/3} \ - \ a^{n/3}}{x \ - \ a } \ = \ \frac{{x^{n/3}} \ - \ a^{n/3}}{(x^{1/3})^3 \ - \ (a^{1/3})^3 } $$

$$ = \ \frac{{x^{n/3}} \ - \ a^{n/3}}{(x^{1/3} \ - \ a^{1/3}) \ \left[ (x^{1/3})^2 \ + \ (x^{1/3})(a^{1/3}) \ + \ (a^{1/3})^2 \right] } \ \ , $$

applying the "difference of two cubes".

Now the issue becomes what positive integer $ \ n \ $ is. If $ \ n = 1 \ , $ we'll just have

$$ \frac{1}{x^{2/3} \ + \ x^{1/3} \ a^{1/3} \ + \ a^{2/3} } \ \ , $$

after canceling numerator against denominator factor. For $ \ n = 2 \ , $ we'd have

$$ \frac{(x^{1/3} \ - \ a^{1/3}) \ ( x^{1/3} \ + \ a^{1/3})}{(x^{1/3} \ - \ a^{1/3}) \ \left[ (x^{1/3})^2 \ + \ (x^{1/3})(a^{1/3}) \ + \ (a^{1/3})^2 \right] } \ \ , $$

$$ = \ \frac{ x^{1/3} \ + \ a^{1/3}}{ x^{2/3} \ + \ x^{1/3} \ a^{1/3} \ + \ a^{2/3} } \ \ , $$

using "difference of two squares" in the numerator first. You should always be able to cancel the "difference of cube roots" factors between the numerator and denominator, though the expression may cease to be so tidy for large values of $ \ n \ . $

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