# 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.

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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 $C$.

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The "rule" on how to choose the numerator in a decomposition can be understood as writing the most general polynomial whose degree is one less than that of the denominator. Why one less? Because if the degree of the numerator was at least the degree of the denominator, we could perform long division and get a polynomial on the LHS, which we clearly don't want given that the RHS is strictly rational. – Austin Mohr Nov 17 '11 at 21:53

$x^2+x+1=(x-a)(x-\bar{a})$ 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 $(2)$.

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I think the method of partial fraction decomposition is precisely where OP needs help. – Austin Mohr Nov 17 '11 at 21:13
@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