# Thread: a standard identity a^n - b^n = (a-b)[a^(n-1)+a^(n-2)b+a^(n-3)b^2+...+b^(n-1)]

1. ## a standard identity a^n - b^n = (a-b)[a^(n-1)+a^(n-2)b+a^(n-3)b^2+...+b^(n-1)]

Why a^n - b^n = (a-b)[a^(n-1)+a^(n-2)b+a^(n-3)b^2+...+b^(n-1)]?
How does one come up with this identity?

2. ## Re: a standard identity a^n - b^n = (a-b)[a^(n-1)+a^(n-2)b+a^(n-3)b^2+...+b^(n-1)]

Are you asking how one can come up with this equation or how to prove it? To prove it, multiply through the right-hand side and almost all terms cancel out.

3. ## Re: a standard identity a^n - b^n = (a-b)[a^(n-1)+a^(n-2)b+a^(n-3)b^2+...+b^(n-1)]

I know how to prove it. I want to know how one can come up with this identity. Can you tell me?

4. ## Re: a standard identity a^n - b^n = (a-b)[a^(n-1)+a^(n-2)b+a^(n-3)b^2+...+b^(n-1)]

Originally Posted by startanewww
I know how to prove it. I want to know how one can come up with this identity. Can you tell me?
You could always long divide \displaystyle \displaystyle \begin{align*} a^n - b^n \end{align*} by \displaystyle \displaystyle \begin{align*} a - b \end{align*} and see what you get...

5. ## Re: a standard identity a^n - b^n = (a-b)[a^(n-1)+a^(n-2)b+a^(n-3)b^2+...+b^(n-1)]

well, most people know this one:

a2 - b2 = (a + b)(a - b)

and this one, too, is fairly common:

a3 - b3 = (a2 + ab + b2)(a - b).

it's not too hard to see that:

a4 - b4 = (a2 + b2)(a + b)(a - b) = (a3 + a2b + ab2 + b3)(a - b)

do you see the pattern...? once you are at this point, it's natural to conjecture that:

$\displaystyle a^n - b^n = (a - b)\left(\sum_{k = 0}^{n-1} a^{n-k-1}b^k \right)$, which you say you can prove.