The partial fraction decomposition of $\dfrac{x}{x^3-1}$

Question

I was trying to decompose $\dfrac{x}{x^3-1}$ into Partial Fractions. I tried the following: $$\dfrac{x}{(x-1)(x^2+x+1)}=\dfrac{A}{(x-1)}+\dfrac{B}{(x^2+x+1)}$$ $$\Longrightarrow A(x^2+x+1)+B(x-1)=x$$ Putting $x=1,$ $$A(1+1+1)+B(1-1)=1\Rightarrow A=\dfrac{1}{3}$$ $$\Rightarrow\dfrac{1}{3}(x^2+x+1) +B(x-1)=x$$ $$\Rightarrow (x^2+x+1)+3Bx-3B=x$$ $$x^2+(3B+1)x +(1-3B)=x$$ At this point I gave up thinking I must have done something wrong since there is no $x^2$ on the RHS while it is there in the LHS.

I would be truly grateful if somebody would be so kind to help me understand my mistake. Many, many thanks in advance!

Answer 1

In a partial fraction decomposition, the numerator of a term with a quadratic denominator (or a power of a quadratic in the denominator) should have the form $$Bx+C,$$ and not just a constant. Tracing through your argument, we see we pick up the algebra at the antepenultimate display equation: $$\frac{1}{3}(x^2 + x + 1) = (B x + C)(x - 1).$$ So, we can see that this correction introduces a new quadratic term $Bx^2$ (with an unknown coefficient), which fixes precisely the problem you've observed with the given method.

Answer 2

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

$\frac{x}{(x-1)(x^2+x+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+x+1}$

$x=A(x^2+x+1)+(Bx+C)(x-1)$

Let $x=1$

$1=A(1^2+1+1)$

$1=3A$

$A=\frac{1}{3}$

So, $x=\frac{1}{3}(x^2+x+1)+(Bx+C)(x-1)$

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

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

Compare the coefficient,

$0x^2=(\frac{1}{3}+B)x^2$

$\frac{1}{3}+B=0$, $B=-\frac{1}{3}$

$x=(\frac{1}{3}-B+C)x$

$\frac{1}{3}-B+C=1$

$\frac{1}{3}-(-\frac{1}{3})+C=1$

$C=\frac{1}{3}$

Therefore,

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

Simplify it and get $\frac{1}{3(x-1)}+\frac{1-x}{3(x^2+x+1)}$