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Homework Help: Expanding x^n-a^n without Binomial Theroem ?

  1. Jun 30, 2010 #1
    1. The problem statement, all variables and given/known data

    33ddulu.jpg

    This is the given Theorem in my book, everything seems fine except that I cannot figure how they expanded (xn - an)

    2. Relevant equations

    The Binomial Theorem

    3. The attempt at a solution

    According to me (xn - an) = {[(x+a)-a]n - an} and expanding it would yield terms containing the nC0,nC1 etc. but they haven't shown anything like this where did all this disappear ? Plus I know that (x-a) would come out common and get cancel by (x-a) in the denominator.
     
  2. jcsd
  3. Jun 30, 2010 #2

    vela

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    The very first phrase in the proof tells you: Dividing (xn-an) by (x-a).
     
  4. Jun 30, 2010 #3
    ^ That went over my head :redface:
    I mean thats the question I am asking, how do I divide them ?
     
  5. Jun 30, 2010 #4

    vela

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  6. Jun 30, 2010 #5

    HallsofIvy

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    Part of the confusion may be that they are NOT expanding- they are factoring which is, basically, the opposite of "expanding".
    You probably already know the second degree version of that: [itex]x^2- y^2= (x- y)(x+ y)[/itex].

    The third degree version is [itex]x^3- y^3= (x- y)(x^2+ xy+ y^2)[/itex].

    In general [itex]x^n- y^n= x^{n-1}+ x^{n-2}y+ x^{n-3}y^2+ \cdot\cdot\cdot+ x^2y^{n-3}+ xy^{n-2}+ y^{n-1}[/itex]
     
  7. Jun 30, 2010 #6
    Thanks very much vela & HallsofIvy, finally I get the hang of this thing, how clumsy of me not to think of it.
     
  8. Oct 13, 2011 #7
    There is small change required in the formula you mentioned.

    (x^n - a^n) must be expanded in general as below.

    (x^n - a^n) = (x - a) ( Rest of what you mentioned above after = sign).

    So, now, (x - a) can be cancelled with the denominator in the problem raised above.

    Suggest me if I am wrong.
     
  9. Oct 13, 2011 #8

    Mark44

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    As already noted in this thread, xn - an is NOT being expanded; it is being factored.
     
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