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Seeing the answer by Andre Nicolas to this question, it got me thinking: is there a way one can derive the formula for $a^n-b^n$ in a more general setting than an euclidian domain, or at least the fact that $a^n-b^n$ is divisible by $a-b$? It seems like all there's needed for it to work is that the ring be commutative, because then: $$ (a-b)(\sum\limits_{k=1}^n a^{n-k}b^{k-1}) = \sum\limits_{k=1}^n a^{n-k+1}b^{k-1} - \sum\limits_{k=1}^n a^{n-k} b^k = a^n - b^n, $$ commutativity being used for the second equality.

Of course, one can do the above check, but it's not nearly as elegant as using the existence of a division algorithm for example, as in the answer I mentioned. So I'm asking if there's a way to at least deduce the divisibility by $a-b$ while also refusing to guess what the formula is.

user26857
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rosecabbage
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  • In rings where $ab\ne ba$, the (left or right) divisibility of $a^n-b^n$ by $a-b$ will in general not hold. So I wonder what way of refusing to guess the result you expect beyond the referenced answer? -- Btw, the division algorithm is used in $R[b][X]$ not in $R$ – Hagen von Eitzen Jan 26 '21 at 13:10
  • Yes, I can see that the ring needs to be commutative if the result is to hold. Now I can see that the answer I mentioned would only work if $R[b]X$ is an euclidian domain, and that means $R[b]$ must be a field. What I'm asking then is, can we do something similarly "clever" to derive the formula if $R[b]$ is not a field, just an arbitrary commutative ring? – rosecabbage Jan 26 '21 at 13:24

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