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.
Since $\displaystyle 1-x^3=(1-x)(1+x+x^2)$,
$\displaystyle \frac{1}{(1-x)(1-x^3)}=\frac{1}{(1-x)^2(1+x+x^2)}=\frac{1}{(x-1)^2(x^2+x+1)}$, and $\displaystyle x^2+x+1$ is not factorable, so you set
$\displaystyle \frac{1}{(x-1)^2(x^2+x+1)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{Cx+D}{x^2+x+1}$
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.
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