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I am trying to factor $r^8-1$ after having just learned about difference and sum of cubes.

I don't know what to do next though as this doesn't seem like a difference of cubes.

How do I factor this in the intended way? I suppose I could just do brute force polynomial division but I don't think that is how it was intended to be factored.

nitrous2
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    $r^8-1 = (r-1)(r+1)(r^2+1)(r^4+1)$ – r9m Mar 28 '14 at 17:00
  • You're right, it is not a difference of cubes because $r^8$ is not a cube. It is, however, a difference of squares, and you probably know already how to handle that. – MJD Mar 28 '14 at 17:27

3 Answers3

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Use the fact that $a^2-b^2=(a-b)(a+b)$.

$r^8-1$

$=(r^4-1)(r^4+1)$

$=(r^2-1)(r^2+1)(r^4+1)$

$=(r-1)(r+1)(r^2+1)(r^4+1)$

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Hint: note first that $r^8-1 = (r^4-1)(r^4+1)$. (Whenever you see $x^{2n}-1$, you should immediately think to factor it as $(x^n-1)(x^n+1)$ since it is a difference of squares.) What can you do from here?

Cameron Williams
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Let $\alpha = \frac {1+i}{\sqrt 2}$, then $$r^8-1=(r-1)(r+1)(r^2+1)(r^4+1) =$$ $$(r-1)(r+1)(r+i)(r-i)(r^2+i)(r^2-i)=$$ $$(r-1)(r+1)(r+i)(r-i)(r+\alpha)(r-\alpha)(r+\bar\alpha)(r-\bar\alpha)$$

As a factorisation into linear factors over the complex numbers. Now noting that $\alpha\bar \alpha=1$ and $\alpha+\bar \alpha=\sqrt 2$ we also have this equal to:

$$(r-1)(r+1)(r^2+1)(r^2-\sqrt2r+1)(r^2+\sqrt 2r+1)$$ As linear and quadratic factors over $\mathbb R$.

So it depends a bit what kind of factorisation you want (and there are neater ways of writing some of these in terms of trigonometric functions or complex exponentials which generalise).

Mark Bennet
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  • I'm not sure if he wanted an irreducible factorization. It probably would have been enough to stop at $(r-1)(r+1)(r^2+1)(r^4+1)$, but kudos to you for going further than that. – recursive recursion Mar 28 '14 at 17:30
  • @recursiverecursion I wasn't sure either, in fact I thought probably not. But I also thought it might be a bit interesting - for anyone who knows about $i$ to be reminded that this can be used, and then I just kept going. I enjoyed writing it. – Mark Bennet Mar 28 '14 at 17:33