Answer
$\frac{1}{x^{2n}-1}$.
Work Step by Step
The given expression is
$\Rightarrow \frac{1}{x^n-1}-\frac{1}{x^n+1}-\frac{1}{x^{2n}-1}$
Let $x^n=t$.
$\Rightarrow \frac{1}{t-1}-\frac{1}{t+1}-\frac{1}{t^{2}-1}$
Factor $t^2-1$.
$=t^2-1^2$
Use the special formula $a^2-b^2=(a+b)(a-b)$.
$=(t+1)(t-1)$
$\Rightarrow \frac{1}{t-1}-\frac{1}{t+1}-\frac{1}{(t+1)(t-1)}$
The LCD of the denominators is $(t+1)(t-1)$.
Multiply the numerators and the denominators
$\Rightarrow \frac{(t+1)}{(t+1)(t-1)}-\frac{(t-1)}{(t+1)(t-1)}-\frac{1}{(t+1)(t-1)}$
Add the numerators because all denominators are equal.
$\Rightarrow \frac{(t+1)-(t-1)-1}{(t+1)(t-1)}$
Simplify.
$\Rightarrow \frac{t+1-t+1-1}{(t+1)(t-1)}$
Add like terms.
$\Rightarrow \frac{1}{(t+1)(t-1)}$
Use the special formula $(a+b)(a-b)=a^2-b^2$.
$\Rightarrow \frac{1}{(t^2-1^2)}$
Back substitute $t=x^n$.
$\Rightarrow \frac{1}{x^{2n}-1}$.
