# Math Help - Partial fraction decomposition

1. ## Partial fraction decomposition

How would I find the partial fraction decomposition for 1/((1-x)(1-x^3))?

2. Originally Posted by thefrog
How would I find the partial fraction decomposition for 1/((1-x)(1-x^3))?
Factorise 1 - x^3 using the difference of two cubes formula. Now see that the denominator has a two-fold repeated linear factor and an irreducible queadratic factor. Now write down the partial fractions corresponding to each. Show all your work and say where you're stuck if you need more help.

3. Since $1-x^3=(1-x)(1+x+x^2)$,

$\frac{1}{(1-x)(1-x^3)}=\frac{1}{(1-x)^2(1+x+x^2)}=\frac{1}{(x-1)^2(x^2+x+1)}$, and $x^2+x+1$ is not factorable, so you set

$\frac{1}{(x-1)^2(x^2+x+1)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{Cx+D}{x^2+x+1}$

When there's a squared factor in the denominator, you get fractions with squared and unsquared denominator. When there's a non-factorable quadratic in the denominator, you get a linear term in the corresponding numerator.

- Hollywood

Edit: Oops - I didn't realize Mr. Fantastic had already posted. So you get two of the same answer.

4. So you're saying I should have believed what Mathematica gave me, then?Thanks, but why is this placed in the precalculus forums? Partial Fractions are not precalculus; I learned calculus well before I first met partial fractions.

5. I also learned partial fractions when I learned how to integrate rational functions (about 2/3 of the way through the first year of calculus).

But it can be understood with pre-calculus knowledge, so I don't think the thread has to be moved necessarily.

- Hollywood

6. Originally Posted by thefrog
So you're saying I should have believed what Mathematica gave me, then?Thanks, but why is this placed in the precalculus forums? Partial Fractions are not precalculus; I learned calculus well before I first met partial fractions.
The rule of thumb is very simple: If the solution of a question does not require calculus then clearly the question does not belong in the Calculus subforum.