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Math Help - 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. #1
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    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.
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    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.
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    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?
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    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 View Post
    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 \begin{align*} a^n - b^n \end{align*} by \displaystyle \begin{align*} a - b \end{align*} and see what you get...
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    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:

    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.
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