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