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X^n-y^n proof

  1. Jun 21, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove [tex]x^n-y^n= (x-y)(x^{n-1}+x^{n-2}y+ \cdots + xy^{n-2}+y^{n-1})[/tex].



    3. The attempt at a solution

    [tex]x^n-y^n = \\
    =x^n -y^n + x^{n-1}y - x^{n-1}y + x^{n-2}y^2 - x^{n-2}y^2 + y^{n-1}x - y^{n-1}x + y^{n-2}x^2 - y^{n-2}x^2
    [/tex] Adding inverses

    [tex]
    = x^n - x^{n-1}y -y^n + y^{n-1}x + x^{n-1}y - x^{n-2}y^2 - y^{n-1}x + y^{n-2}x^2 + (x^{n-2}y^2 - y^{n-2}x^2)[/tex]
    Rearranging inverse in order to be factored.


    I cannot get rid of the last term in parentheses. I realize that all the combined terms in the last step can be factored with something and left with (x-y) which is a step closer to the proof. The last term is the bane of this problem. Any insight please?

    Sorry if the algebra looks obfuscating; it gives me headaches.
     
  2. jcsd
  3. Jun 21, 2009 #2

    D H

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    Why such a haphazard expansion?

    You need to account for all of the terms. In particular, you appear to be forgetting that the ellipsis encompasses a lot of terms (everything from xn-3y2 to x2yn-3).
     
  4. Jun 21, 2009 #3
    You do not need to get rid of the last term in parentheses. You are on the right way, just you need to factor the negative terms (with minus before them), and the positive terms (with plus before them). Group the negative and positive terms in separate brackets, and factor y for the negative ones, and x for the positive ones.

    Regards.
     
  5. Jun 21, 2009 #4

    Borek

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    Won't it be easier to multiply and cancel the RHS?
     
  6. Jun 21, 2009 #5
    "In particular, you appear to be forgetting that the ellipsis encompasses a lot of terms (everything from xn-3y2 to x2yn-3). "

    Yeah I forgot to include them.


    "You are on the right way, just you need to factor the negative terms (with minus before them), and the positive terms (with plus before them). Group the negative and positive terms in separate brackets, and factor y for the negative ones, and x for the positive ones."

    In the last step, I arranged them to be factored but not by the method you proposed. I do not see how your method would just by factoring y and x alone. Unless you mean making a lot of individual brackets--what I originally intended, I do not see that would work otherwise.

    This is what i meant to do :

    [tex]= (x^n - x^{n-1}y)+( -y^n + y^{n-1}x )+ (x^{n-1}y - x^{n-2}y^2) + ( - y^{n-1}x + y^{n-2}x^2) + (x^{n-2}y^2 - y^{n-2}x^2)
    [/tex]
    Then I would factor the largest common factor thus leaving (x-y) or (y-x) in each case. The problem arises in the last parentheses because I cannot factor anything from there.

    "Won't it be easier to multiply and cancel the RHS?"

    What is RHS?
     
  7. Jun 21, 2009 #6

    dx

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    RHS = Right Hand Side
     
  8. Jun 21, 2009 #7
    That defeats the entire purpose of the problem, if I understand you correctly.
     
  9. Jun 21, 2009 #8

    dx

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    No, it's a valid way to prove it.
     
  10. Jun 21, 2009 #9
    I am trying to get to that result. I think multiplying the RHS and cancelling with the terms on the LHS just tests if they are equal.
     
  11. Jun 21, 2009 #10

    dx

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    You don't cancel anything with the LHS. Just expandthe RHS, and show that it is equal to the LHS.
     
  12. Jun 21, 2009 #11
    But that is not what I am supposed to do; I am supposed to work it out and arrive at that result that they set equal to. My(and yours) efforts are supposed to [tex]= (x-y)(x^{n-1}+x^{n-2}y+ \cdots + xy^{n-2}+y^{n-1}) [/tex].
     
  13. Jun 21, 2009 #12

    dx

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    You are supposed to prove that

    [tex] x^n-y^n= (x-y)(x^{n-1}+x^{n-2}y+ \cdots + xy^{n-2}+y^{n-1}) [/tex]

    is an identity, correct?

    If so, then expanding the RHS and showing that it is equal to the LHS is a valid way to do that.
     
  14. Jun 21, 2009 #13
    Any alternatives beside this?
     
  15. Jun 21, 2009 #14

    Borek

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    Any way of showing that LHS equals RHS will do. Your original approach was correct as well - I just think dealing with RHS is easier.
     
  16. Jun 21, 2009 #15

    Fredrik

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    Razored, it's very close to 100% certain that the person who gave you this problem to solve intended you to do it the way Borek and dx are suggesting. You do know that A=B if and only if B=A, right? You don't have to start with the thing that's on the left. You can also start with the thing that's on the right and show that it's equal to what's on the left.
     
  17. Jun 21, 2009 #16

    D H

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    Why do you want an alternative? The whole point of this exercise is to learn how to multiply polynomials.

    There are n terms in xn-1y+xn-2y2+...+x2yn-2+xyn-1. The product
    (x-y)(xn-1y+xn-2y2+...+x2yn-2+xyn-1) thus has 2n terms. You need to show that all but two of them cancel.
     
  18. Jun 21, 2009 #17
    Yeah, I should have multiplied out the RHS. I will leave it at that. I guess the author did intend it as a practice of multiplying polys.

    Thanks.
     
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