Converting multiplying fractions to sum of fractions

Question

I have the next fraction: $$\frac{1}{x^3-1}.$$

I want to convert it to sum of fractions (meaning $1/(a+b)$).

So I changed it to: $$\frac{1}{(x-1)(x^2+x+1)}.$$ but now I dont know the next step. Any idea?

Thanks.

Answer 1

The process here is partial fraction decomposition. The first step, which you've kindly done already, is to factor the denominator completely. Now, note that if we had a sum of the form $$ \frac{\text{something}}{x-1} + \frac{\text{something}}{x^2 + x + 1} $$ then we could multiply the left fraction by $\frac{x^2 + x + 1}{x^2 + x + 1}$ and the right fraction by $\frac{x-1}{x-1}$ and then the denominators would both match the original one, so they might just add up to our original fraction! Let's try to find such a decomposition.

The way we can do this is pretty much to just write the above equation, but a little more specifically. The rule is that the $\text{something}$ that goes over a linear factor (e.g. $x-1$) is a single variable, say $A$; and the $\text{something}$ that goes over a quadratic factor (e.g. $x^2 + x + 1$) is linear, that is it has the form $Bx + C$. So here is our equation: $$ \frac{\text{A}}{x-1} + \frac{\text{Bx+C}}{x^2 + x + 1} = \frac{1}{(x-1)(x^2 + x + 1)} $$ We can now perform the multiplication suggested above to get the numerator on the left side in terms of $A$, $B$, and $C$, and the denominators equal. The denominators cancel each other then, so we know this numerator must equal $1$, and more clearly it must equal $0x^2 + 0x + 1$ so we can use the coefficients of the terms in the numerator to find a system of equations (the $x^2$ terms must add to zero, the $x$ terms must add to zero, etc.) and solve for $A$, $B$, and $C$.

Answer 2

$x^2+x+1=(x-a)(x-\bar{a})$ where $a=\exp(\frac{2\pi i}{3})=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$, so $$ \frac{1}{x^3-1}=\frac{1}{(x-1)(x-a)(x-\bar{a})}\tag{1} $$ and then you can use partial fractions on $(1)$ to get $$ \frac{1}{x^3-1}=\frac{1}{3}\left(\frac{1}{x-1}+\frac{a}{x-a}+\frac{\bar{a}}{x-\bar{a}}\right)\tag{2} $$ Partial Fractions (Heaviside Method):

Suppose we wish to write $$ \frac{A}{x-1}+\frac{B}{x-a}+\frac{C}{x-\bar{a}}+\frac{1}{(x-1)(x-a)(x-\bar{a})}\tag{3} $$ To compute $A$, multiply both sides by $x-1$ and set $x=1$: $$ \begin{align} A &=\frac{1}{(1-a)(1-\bar{a})}\\ &=\frac{1}{3}\tag{3a} \end{align} $$ To compute $B$, multiply both sides by $x-a$ and set $x=a$: $$ \begin{align} B &=\frac{1}{(a-1)(a-\bar{a})}\\ &=\frac{a}{3}\tag{3b} \end{align} $$ To compute $C$, multiply both sides by $x-\bar{a}$ and set $x=\bar{a}$: $$ \begin{align} C &=\frac{1}{(\bar{a}-1)(\bar{a}-a)}\\ &=\frac{\bar{a}}{3}\tag{3c} \end{align} $$ Collecting equations $(3)$, yields $(2)$.