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Prof = Wrong?

  1. Sep 20, 2007 #1
    1. The problem statement, all variables and given/known data
    lim |x-1|/x^3-1
    x-->1+



    2. Relevant equations



    3. The attempt at a solution

    This is from a "practise test" and the prof wrote his solution on it... I'm having a hard time figuring out why (x^3-1) = (x-1)(x^2+x+1) which is what he used to find the limit.... im fine at figureing out limits but I just have no idea how he figures those two equations are equal because when i multiply through i get (x^3-2x^2+2x-1)
     
  2. jcsd
  3. Sep 20, 2007 #2

    cristo

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    (x-1)(x^2+x+1)=x^3+x^2+x-x^2-x-1=x^3-1 where the first three terms are obtained by multiplying the first term in the first bracket through with the right bracket, and the last three terms are obtained by multiplying the second term in the first bracket through with the right bracket.
     
  4. Sep 20, 2007 #3

    D H

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    You're professor is right. Try multiplying it out again. If you get the same (wrong) answer again, show your work, and we'll help show you how to get to the correct answer.

    Edit:
    Well forget that. Cristo just did it for you.
     
  5. Sep 20, 2007 #4
    Lol, thank you on my paper i have it written down as (x-1)(x^2-x+1)...

    But I am still wondering how you get from (x^3-1) TO (x-1)(x^2+x+1)
    I have tried to factor by grouping but that got me nowhere...
     
    Last edited: Sep 20, 2007
  6. Sep 20, 2007 #5

    D H

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    I don't remember that equation specifically. I do remember that
    [tex]x^n-1 = (x-1)(x^{n-1}+x^{n-2}+\cdots+x+1)[/itex]
    Equations of this form come up a lot. It is worth knowing.

    If not, you have to factor. In general, yech. There is a general rule for solving cubics. I use it so infrequently I have to look it up every time I use it. There is a general rule for solving quartics, which I never use. There is no general rule for higher order polynomials. On the other hand, I run into things like [itex]1/(1+x)[/itex], [itex]1/(x-1)[/itex], and [itex](x^n-1)/(x-1)[/itex] all the time.
     
  7. Sep 20, 2007 #6
    (x^3-1)=(x^3+x^2-x^2+x-x-1)
    =(x^3+x^2-x^2)+(+x-x-1)
    =-1(-x^3-x^2+x^2)-1(-x+x)
    =-1(i dont know....
     
  8. Sep 20, 2007 #7

    cristo

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    You can spot that x-1 is a factor and then use polynomial division. I don't think I've ever used the cubic formula!
     
  9. Sep 20, 2007 #8
    is it called the binomial theorm?
     
  10. Sep 21, 2007 #9
    There is a fairly well known equation for difference of cubes:

    [tex](a^3-b^3)=(a-b)(a^2+ab+b^2)[/tex]

    It's fairly easy to prove, once you just multiply everything out. There is also an analogue for addition of cubes which I do not provide for the reader.
     
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