# 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
Commented Feb 21, 2013 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
Commented Feb 21, 2013 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. Commented Feb 21, 2013 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
Commented Feb 21, 2013 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. Commented Feb 21, 2013 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... Commented Feb 21, 2013 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
Commented Feb 21, 2013 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). Commented Feb 21, 2013 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. Commented Dec 26, 2015 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. Commented Dec 26, 2015 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. Commented Dec 26, 2015 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. Commented Nov 27, 2015 at 16:52
• Multiplying out the right hand side will surely show that the factorization is correct, but it would provide no motivation for where it came from in the first place, which geometric series does provide Commented Apr 16 at 9:30