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How does one arrive at this ?

  1. Aug 9, 2009 #1
    How does one arrive at this ???

    I was wondering how does someone get from factorising
    :{a^(n)-b^(n)}

    to
    {a-b}{a^(n-1)-a^(n-2)(b)+a^(n-3)b^(2)----+b^(n-1)}

    thanks
     
  2. jcsd
  3. Aug 9, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi rbnphlp! Welcome to PF! :smile:

    (try using the X2 tag just above the Reply box :wink:)

    Just multiply the long one by a on one line, and by b on another line, and subtract. :smile:
     
  4. Aug 9, 2009 #3
    Re: Welcome to PF!

    Sorry , but my original question is how do I arrive at the long one from (an-bn),
     
  5. Aug 9, 2009 #4

    tiny-tim

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    ah!

    use long division, exactly as you would for decimals. :smile:
     
  6. Aug 9, 2009 #5

    jgens

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    Re: How does one arrive at this ???

    Clearly [itex]a=b[/itex] is a solution to [itex]a^n-b^n[/itex]. To factor out the [itex]a-b[/itex] term, we use polynomial long division (or synthetic polynomial division) to find that [itex]\frac{a^n-b^n}{a-b} = a^{n-1} + \dots + b^{n-1}[/itex].
     
  7. Aug 9, 2009 #6
    Re: How does one arrive at this ???

    Although polynomial long division can arrive at the desired quotient, it is usually easier to demonstrate the equality by just distributing the left factor on the right-hand side across its cofactor:

    [tex](a-b)\cdot(a^{n-1} + a^{n-2} b + \cdots + ab^{n-2} + b^{n-1})
    = (a^n + a^{n-1}b + \cdots + ab^{n-1}) - (a^{n-1}b + a^{n-2}b^2 + \cdots + b^n)[/tex]

    and then simplifying.

    --Elucidus
     
  8. Aug 10, 2009 #7
    Re: How does one arrive at this ???

    oh got it..thankyou
     
  9. Aug 10, 2009 #8
    Re: How does one arrive at this ???

    thanks evryone
     
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