Answer
$\dfrac{1}{x^{n}-1}-\dfrac{1}{x^{n}+1}-\dfrac{1}{x^{2n}-1}=\dfrac{1}{x^{2n}-1}$
Work Step by Step
$\dfrac{1}{x^{n}-1}-\dfrac{1}{x^{n}+1}-\dfrac{1}{x^{2n}-1}$
Evaluate the subtraction of the first two terms and simplify:
$\dfrac{1}{x^{n}-1}-\dfrac{1}{x^{n}+1}-\dfrac{1}{x^{2n}-1}=...$
$...=\Big(\dfrac{1}{x^{n}-1}-\dfrac{1}{x^{n}+1}\Big)-\dfrac{1}{x^{2n}-1}=...$
$...=\dfrac{x^{n}+1-x^{n}+1}{(x^{n}-1)(x^{n}+1)}-\dfrac{1}{x^{2n}-1}=...$
$...=\dfrac{2}{(x^{n}-1)(x^{n}+1)}-\dfrac{1}{x^{2n}-1}=...$
The product on the denominator of the first fraction is the result of factoring a difference of squares. Evaluate this product:
$...=\dfrac{2}{(x^{n})^{2}-1^{2}}-\dfrac{1}{x^{2n}-1}=\dfrac{2}{x^{2n}-1}-\dfrac{1}{x^{2n}-1}=...$
Both terms now have the same denominator. Evaluate this last subtraction:
$...=\dfrac{2-1}{x^{2n}-1}=\dfrac{1}{x^{2n}-1}$
