How to factor $a^n - b^n$?

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?

• 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
• @asdf: There is a process called "homogenization" of polynomials that it is nice to be aware of. If $p(x)=c_nx^n+\cdots+c_1x+c_0$ is any polynomial, then $$b^np(a/b)=c_na^n+c_{n-1}a^{n-1}b+c_{n-2}a^{n-2}b^2+\cdots+c_2a^2b^{n-2}+c_1a b^{n-1}+c_0b^n.$$ Ultimately, it may boil down to pattern recognition: the monomials $a^kb^{n-k}$ (as $k$ varies) may be rewritten as $(a/b)^kb^n$, and $b^n$ does not vary with $k$ while $(a/b)^k$ is easier to work with. – anon Feb 21 '13 at 1:26
• 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

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

• 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

I see the answer is accepted. But for future reference, another proof would be

Let $$p(x)=x^n-a^n$$. Clearly, $$x=a$$ is a solution. This means $$x-a$$ is a factor of $$x^n-a^n.$$

It is just a matter of simple polynomial division aafter that and so dividing $$x^n-a^n$$ by $$x-a$$ gives us $$x^{n-1} + ax^{n-2} +\cdots + a^{n-1}$$

So, $$x^n-a^n=(x-a)(x^{n-1} + ax^{n-2} +\cdots + a^{n-1}).$$

Replace $$x$$ and $$a$$ with $$a$$ and $$b$$.

• That proves $a-b$ is one factor, but gives no clue to the other one. – vonbrand Dec 26 '15 at 2:05
• Its just simple polynomial division after that. I thought people would figure that out on their own but I will editit now. Thanks. – Skawang Dec 26 '15 at 15:38
• I am sorry but I don't know how exactly to write polynomial long division in latex. I would appreciate it if someone did it for me. – Skawang Dec 26 '15 at 15:45

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

• This seems to me to be the best explanation of that factorization. I would have said that calling in the formula for a finite geometric series was overkill. – Lubin Nov 27 '15 at 16:52