# Partial Fractions?

1. Apr 10, 2012

### hqjb

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

$\frac{1}{x^2+x+1} = \frac{1-x}{1-x^3}$

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. Apr 10, 2012

### 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?

Last edited: Apr 10, 2012
3. Apr 10, 2012

### hqjb

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.

4. Apr 10, 2012

### Mentallic

If you want to go from the LHS to the RHS of $$\frac{1}{x^2+x+1}\equiv \frac{1-x}{1-x^3}$$ 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 $$\frac{1}{x^2+x+1}\cdot \frac{1-x}{1-x}$$ to get the numerator the way you want it, and simultaneously keeping the expression equal.

5. Apr 10, 2012

### hqjb

Alright thanks!