As the title states. I don't quite understand how this factoring would work even though I'm fairly certain it can be done. Additionally, a follow up would be how does factoring $x^{n-1} - y^{n-1}$ out of $x^n - y^n$ work.
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1Are you familiar with polynomial division? Try dividing $(x^n-y^n)$ by $(x-y)$ and you should see some patterns emerging. – Jam Feb 16 '20 at 04:36
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I haven't done that in awhile! I understand the summation answer below, but is there no conciser answer? – user2793618 Feb 16 '20 at 04:37
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1I doubt it. I don't think the sum simplifies further. – Jam Feb 16 '20 at 04:38
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Ah ok. It does seem to be that way. – user2793618 Feb 16 '20 at 04:38
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What do you mean by a concise answer? – azif00 Feb 16 '20 at 04:39
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6Does this answer your question? Proving $x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1})$ – Kyan Cheung Feb 16 '20 at 04:40
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1@user2793618, that is the concise answer to this. It is a geometric series, but getting a closed form for that will result in the expression in the question. So there's no use and the sum is the only possible simplification. – superposition Feb 16 '20 at 04:41
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I guess I was looking for a clean way to express it, i.e, no summations or some division trick, but this seems to be the best answer. – user2793618 Feb 16 '20 at 04:41
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What makes you imagine it's possible to factor $x^{n-1}-y^{n-1}$ out of $x^n-y^n$? – David C. Ullrich Feb 16 '20 at 11:05
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$$\begin{align} x^n - y^n &= (x-y)(x^{n-1} + x^{n-2}y + \cdots + xy^{n-2} + y^{n-1}) \\ &= (x-y)\sum_{k=0}^{n-1} x^{n-1-k}y^k \end{align}$$ generalizing the identity $x^3 - y^3 = (x-y)(x^2 + xy + y^2)$.
azif00
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Also not much but a side-comment to this. For $n$ being even. We have $$x^n-y^n=\left(x-y\right)\left(x+y\right)\sum_{k=0}^{n/2-1}x^{\left(n-2-2k\right)}y^{2k}$$ For example, $$x^6-y^6=\left(x-y\right)\left(x+y\right)\left(x^{4}+x^{2}y^{2}+y^{4}\right)$$
