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Spivak: Is his how to approach these proofs?

  1. Sep 1, 2011 #1
    Hello all :smile:

    I have started the problem set for Chapter one (basic properties of numbers) in Spivak's Calculus (self study). I think I am doing these right, but I have some questions.

    As a solid example, problem 1-(iv) says to prove the following:

    [tex]x^3 - y^3 = (x-y)(x^2+xy+y^2)\qquad(1)[/tex]

    My approach was to add and subtract terms to the left hand side of (1) until it could be factored into the desired form:

    x^3 - y^3 &= x^3 - y^3 + (x^2y - x^2y) + (xy^2 - xy^2) \\
    &= x^3 +x^2y+xy^2 - y^3 - x^2y - xy^2 \\
    &=x(x^2+xy+y^2)-y(x^2+xy+y^2) \\
    &= (x-y)(x^2+xy+y^2)

    Now this seems correct to me, but I feel a little guilty because I only knew what to add and subtract to the LHS of (1) because I knew what I was trying to achieve- the RHS of (1).

    My second question is similar, but requires some clarifying of Spivak's intent. In problem 1-(vi) he says: Prove the following:

    [tex]x^3 + y^3 = (x+y)(x^2-xy+y^2)\qquad(2)[/tex]

    Then he says, "There is a particularly easy way to do this, using (iv), and it will show you how to find a factorization of [itex]x^n+y^n[/itex] whenever n is odd."

    Well, I solved this one the exact same way I did with (iv) above: I used the right hand side of (2) to infer what terms to add/subtract and obtained the solution. However, the fact that I do not see what he means by the quoted text above leads me to belive that there was some other approach. That is, I have not discovered a way to factorize [itex]x^n+y^n[/itex] whenever n is odd in my procedure.

    Any thoughts are appreciated :smile:
  2. jcsd
  3. Sep 1, 2011 #2


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    Certainly, if you know (1), you can get (2) by simply substituting -y in for the y.
  4. Sep 1, 2011 #3
    Hi LC :smile: Yes. I arrived at the solution, but in particular, I am concerned about my approach as I do not see what he means by

  5. Sep 1, 2011 #4


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    OK, I see what you mean now. Apparently you need a more clever approach to (1) that easily generalizes to higher odd exponents. I dunno -- I'll think about it if this football game doesn't keep me too distracted. But don't count on me for this one :frown:
  6. Sep 1, 2011 #5


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    Also, I guess that observing that y = x is a root for odd exponents and factoring (y-x) by long division doesn't count?
  7. Sep 1, 2011 #6
    Maybe he made a typo? He says "this can easily be done using (iv)". Maybe he meant (v). In problem (v) we proved that

    x^n-y^n = (x-y)(x^{n-1} + x^{n-2} y+\dots+xy^{n-2} +y^{n-1} )

    Anyway, no big deal! Get back to your game! :smile:
  8. Sep 2, 2011 #7
    My thoughts:

    Spivak is correct, (vi) can be easily done by (iv) by just substituting -y for y.
    This gives you a hint how to obtain the factorization of [itex]x^n+y^n[/itex]. Indeed, take (v) and substitute -y for y again...
  9. Sep 2, 2011 #8
    I am pretty sure this is what Spivak intended.

    Also note that you don't have to start from the left side for the first one..


    This approach will come in handy for at least one problem in chapter two, but I will let you figure out which one!
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