# Thread: partial fractions cubic base

1. ## partial fractions cubic base

i still cant get the correct answer for this,i simply equated coefficients etc anad got as far as the first two terms in the answer. The question is basically
Q express $\frac{x^5-1}{x^2(x^3+1)}$ in partial fractions ?[/quote] $\frac{x^5-1}{x^5+x^2}$
supposed working,,
$\frac{x^5-1+x^2-x^2}{x^5+x^2}$

$\frac{x^5+x^2}{x^5+x^2}-\frac{1+x^2}{x^5+x^2}$

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

I'd start it that way. You can try and finish it.

In this case:

$1+x^2\equiv(Ax+B)(x^3+1)+(Cx^2+Dx+E)(x^2)$

Since this is an identity, it is true for all values of x (this is a great help in determing the values of A,B,C,D and E!).[/quote] i miust be doing something awfully wrong

2. You can make the last part a bit simpler by factorising $1+x^3$ as $(1+x)(1-x+x^2)$. Then you'll want $\frac{1+x^2}{x^2(1+x)(1-x+x^2)} = \frac{Ax+B}{x^2} + \frac C{1+x} + \frac{Dx+E}{1-x+x^2}$.

3. Hi, thanks for your reply and your method works but i can't help to not understand why it doesnt work when i use the partial fractions, $\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x+1}+\frac{Dx+E }{x^2-x+1}$
Using this u can quickly deduce B=1,,,and C=2/3 by elimiting their factors but then when i put in x=1,,i get -2/3=2(A+D+E)
$(x^4)$,,,E=-1/3
$(x^5)$,,,A+D=-2/3 which then gives me(putting these together):2A+2D=0

I know A must be zero and D cant be zero as the answer to this question is
$1-\frac{1}{x^2}-\frac{2}{3(x+1)}-\frac{1-2x}{3(x^2-x+1)}$
NB previous post was 1 - p(x)/q(x) IF SOMEONE COULD HELP THEYD BE REALLY COOL!!!!XD

4. You seem to be doing it mostly correctly. There must be a simple arithmetic error somewhere. Starting from

$1+x^2\equiv (Ax+B)(1+x^3)+ Cx^2(1-x+x^2) +(Dx+E)x^2(1+x)$,

you get B=1 and C=2/3 (by putting x=0 and x=-1). So the identity becomes

$1+x^2\equiv (Ax+1)(1+x^3)+ \tfrac23x^2(1-x+x^2) +(Dx+E)x^2(1+x)$.

Now compare the coefficients of x on both sides, to get A=0, so the identity becomes

$1+x^2\equiv 1+x^3 + \tfrac23x^2(1-x+x^2) +(Dx+E)x^2(1+x)$.

Next, compare the coefficients of $x^4$ on both sides, to get $0=\tfrac23+D$, so the identity becomes

$1+x^2\equiv 1+x^3 + \tfrac23x^2(1-x+x^2) +(\tfrac23x+E)x^2(1+x)$.

Now you can wrap it up by putting x=1 (say) to get an equation for E.

5. my bad i got it thanks