Converting multiplying fractions to sum of fractions

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.

• Use the method of partial fractions: en.wikipedia.org/wiki/Partial_fractions – user7530 Nov 17 '11 at 20:47
• I don't think you mean "sum of fractions" the way you wrote; you probably mean that you want to write it as a sum of fractions $\frac{1}{a} + \frac{1}{b}$ (otherwise, it's already written in the way you want: $a=x^3$ and $b=-1$). Is this indeed what you want? You may not be able to get it as a sum of "egyptian-like" fractions (numerator equal to $1$). Can you clarify exactly what you mean? – Arturo Magidin Nov 17 '11 at 20:48

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 $C$.
$x^2+x+1=(x-a)(x-\bar{a})$ 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 $(2)$.
• @Austin: Thanks for pointing that out. Since all of my denominators have degree $1$, I have illustrated the Heaviside method for Partial Fractions. – robjohn Nov 18 '11 at 0:52