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?

I really can't figure this out... Please help.

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.

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?

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)]

Quote:

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...

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:

a^{2} - b^{2} = (a + b)(a - b)

and this one, too, is fairly common:

a^{3} - b^{3} = (a^{2} + ab + b^{2})(a - b).

it's not too hard to see that:

a^{4} - b^{4} = (a^{2} + b^{2})(a + b)(a - b) = (a^{3} + a^{2}b + ab^{2} + b^{3})(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.