Answer
$\frac{1}{(x^3-1)(x^2-1)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+1}+\frac{Dx+E}{x^2+x+1}$
Work Step by Step
$\frac{1}{(x^3-1)(x^2-1)}$
For each factor $(ax+b)^n$ of the denominator we add $n$ fractions $\frac{A_k}{(ax+b)^k}$, where $k=1,2,3,\ldots n$.
For each factor $(ax^2+bx+c)^n$, where $ax^2+bc+c$ is irreducible, we add $n$ fractions $\frac{p_k x+q_k}{(ax^2+bx+c)^k}$, where $k=1,2,3,\ldots n$.
Since $x^3-1=(x-1)(x^2+x+1)$ and cannot be factored further, and $x^2-1=(x-1)(x+1)$ and cannot be factored further, we wrtite:
$\frac{1}{(x^3-1)(x^2-1)}=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+1}+\frac{Dx+E}{x^2+x+1}$
