Answer
The required solution is $\frac{1}{{{x}^{2n}}-1}$
Work Step by Step
We have the given rational expression:
$\frac{1}{{{x}^{n}}-1}-\frac{1}{{{x}^{n}}+1}-\frac{1}{{{x}^{2n}}-1}$
Now, simplify the third part of the given expression:
$\begin{align}
& \frac{1}{{{x}^{2n}}-1}=\frac{1}{{{\left( {{x}^{n}} \right)}^{2}}-{{\left( 1 \right)}^{2}}} \\
& =\frac{1}{\left( {{x}^{n}}+1 \right)\left( {{x}^{n}}-1 \right)}
\end{align}$
So, the given expression becomes:
$\frac{1}{{{x}^{n}}-1}-\frac{1}{{{x}^{n}}+1}-\frac{1}{{{x}^{2n}}-1}=\frac{1}{{{x}^{n}}-1}-\frac{1}{{{x}^{n}}+1}-\frac{1}{\left( {{x}^{n}}+1 \right)\left( {{x}^{n}}-1 \right)}$
And simplify the above expression:
$\begin{align}
& \frac{1}{{{x}^{n}}-1}-\frac{1}{{{x}^{n}}+1}-\frac{1}{{{x}^{2n}}-1}=\frac{1}{\left( {{x}^{n}}-1 \right)}\times \frac{\left( {{x}^{n}}+1 \right)}{\left( {{x}^{n}}+1 \right)}-\frac{1}{\left( {{x}^{n}}+1 \right)}\times \frac{\left( {{x}^{n}}-1 \right)}{\left( {{x}^{n}}-1 \right)}-\frac{1}{\left( {{x}^{n}}+1 \right)\left( {{x}^{n}}-1 \right)} \\
& =\frac{\left( {{x}^{n}}+1 \right)}{\left( {{x}^{n}}-1 \right)\left( {{x}^{n}}+1 \right)}-\frac{\left( {{x}^{n}}-1 \right)}{\left( {{x}^{n}}+1 \right)\left( {{x}^{n}}-1 \right)}-\frac{1}{\left( {{x}^{n}}+1 \right)\left( {{x}^{n}}-1 \right)} \\
& =\frac{{{x}^{n}}+1-{{x}^{n}}+1-1}{\left( {{x}^{n}}+1 \right)\left( {{x}^{n}}-1 \right)} \\
& =\frac{1}{\left( {{x}^{n}}+1 \right)\left( {{x}^{n}}-1 \right)}
\end{align}$
Hence, $\frac{1}{{{x}^{n}}-1}-\frac{1}{{{x}^{n}}+1}-\frac{1}{{{x}^{2n}}-1}=$ $\frac{1}{{{x}^{2n}}-1}$.
