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Wikipedia provides a proof, but I don't understand how:

$$a^n - b^n = (a-b)(a^{n-1} + ba^{n-2} +\cdots + b^{n-1})$$

follows from

$$x^{n-1} + x^{n-2} +\cdots + x + 1 = \frac{x^n - 1}{x-1}$$

Could someone explain to me how the summation of the the geometric series explains the factorization?

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Plug $x=a/b$ into the geometric sum formula, then clear denominators (multiply the left by $b^{n-1}$, the right's numerator by $b^n$, and the right's denominator by $b$, then multiply both sides by $a-b$). –  anon Feb 21 '13 at 1:17
    
@anon Thanks anon! Just wondering, but how did you figure that out? It seems like a very complex connection to make- could you please share your train of thought? –  asdf Feb 21 '13 at 1:19
    
But we don't need to obtain the factorization from the $x$ stuff. Just find (or imagine finding) the product $(a-b)(a^{n-1}+a^{n-2}b+\cdots +b^{n-1})$ and observe that almost all the terms in the product cancel. –  André Nicolas Feb 21 '13 at 1:20
    
@AndréNicolas But how could someone just "find" it? How would someone discover such a connection? Is it just the result of randomly trying out different things? Or does it follow from a specific train of thought? –  asdf Feb 21 '13 at 1:23
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Presumably it came from $a^2-b^2=(a-b)(a+b)$, and $a^3-b^3=(a-b)(a^2+ab+b^2)$, and then looking for something similar for higher powers. –  André Nicolas Feb 21 '13 at 1:26

2 Answers 2

up vote 5 down vote accepted

The long parenthesized term is a geometric series with first term $a^{n-1}$ and ratio $\frac ba$ so set $x=\frac ba$

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The two formulae agree better, if you divide the first equation by $b^n$ and then set $x=a/b$ (instead of $x=b/a$). Then simply divide by the first rhs-parenthese... –  Gottfried Helms Feb 21 '13 at 1:13
    
Thanks Ross! One more thing: this might be a strange question, but how could anyone possibly see the intuition between the substitution? I mean how could possibly see the connection between the sum of a geometric series, a fraction substitution and the factorization? It seems like someone connected distinct parts of math and suddenly came upon the factorization! –  asdf Feb 21 '13 at 1:18
    
@asdf: I find it easier to find the factorization by considering what happens with each term. A typical term $a^ib^{n-i-1}$ gets multiplied by $(a-b)$ and makes $a^{i+1}b^{n-i-1}-a^ib^{n-i}$ The first cancels with the second term of the previous one and the second cancels with the first term of the next. All that survives is $a^n$ from the first term (because there is no term before) and $-b^n$ from the last one (because there is no term after). –  Ross Millikan Feb 21 '13 at 1:23

Just multiply out the right hand side, you'll see that all terms except for the left hand side cancel.

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