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Partial Fractions?

  1. Apr 10, 2012 #1
    1. The problem statement, all variables and given/known data
    In one of the workings of a question I couldn't solve(from solution sheet) there was one step I couldn't understand.

    2. Relevant equations

    [itex]\frac{1}{x^2+x+1} = \frac{1-x}{1-x^3}[/itex]

    3. The attempt at a solution

    Tried partial fractions(unfactorable denominator) and could'nt get it, finally got it through trial and error but is there a method where you apply to all such expressions to simplify it?
     
  2. jcsd
  3. Apr 10, 2012 #2

    NascentOxygen

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    Staff: Mentor

    You're wanting to factorize (1-x³). You can see that x=1 solves it, so take out (x-1) as a factor.

    How to do this? Do you know the "long division" method for dividing (1-x³) by (x-1)? This will give you the other factor.

    :: EDIT:: I assumed you were working from the RH expression, wanting to find the LHE. Maybe you weren't. Which do you regard as the "simplest" expression here? :smile:
     
    Last edited: Apr 10, 2012
  4. Apr 10, 2012 #3
    Haha sorry if I didnt put it into context properly. I wanted to obtain the RHE from the LHE, it wasnt a proving question but rather the solution manual skipped steps from LH-> RH and I'm not sure how he did it and if I can replicate it on a similar question without hindsight.

    Edit : It's for Mclaurin's series for I needed it in 1\(1-x) form.
     
  5. Apr 10, 2012 #4

    Mentallic

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    Homework Helper

    If you want to go from the LHS to the RHS of [tex]\frac{1}{x^2+x+1}\equiv \frac{1-x}{1-x^3}[/tex] then what you need is for the numerators to be equivalent, but at the same time you can't change the LHS expression.

    So what you can do is use [tex]\frac{1}{x^2+x+1}\cdot \frac{1-x}{1-x}[/tex] to get the numerator the way you want it, and simultaneously keeping the expression equal.
     
  6. Apr 10, 2012 #5
    Alright thanks!
     
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