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.
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.
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 $(2)$.