<s>[INST] 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.
 [/INST] $x^2+x+1=(x-a)(x-\bar{a})$ 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 $(2)$.
</s>