The expression $\frac{1}{x - 1} - \frac{1}{x + 1} - \frac{2}{x^{2} + 1} - \frac{4}{x^{4} + 1}$ $\frac{1}{{x - 1}} - \frac{1}{{x + 1}} - \frac{2}{{{x^2} + 1}} - \frac{4}{{{x^4} + 1}}$ is equal to = ?

1 $\frac{8}{x^{8} + 1}$ $\frac{8}{{{x^8} + 1}}$
2 $\frac{8}{x^{8} - 1}$ $\frac{8}{{{x^8} - 1}}$
3 $\frac{8}{x^{7} - 1}$ $\frac{8}{{{x^7} - 1}}$
4 $\frac{8}{x^{7} + 1}$ $\frac{8}{{{x^7} + 1}}$

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Given expression, \\ (\frac{1}{x - 1} - \frac{1}{x + 1}) - \frac{2}{x^{2} + 1} - \frac{4}{x^{4} + 1} \\ = [\frac{(x + 1) - (x - 1)}{(x - 1) (x + 1)}] - \frac{2}{x^{2} + 1} - \frac{4}{x^{4} + 1} \\ = (\frac{2}{x^{2} - 1} - \frac{2}{x^{2} + 1}) - \frac{4}{x^{4} + 1} \\ = [\frac{2 (x^{2} + 1) - (x^{2} - 1)}{(x^{2} - 1) (x^{2} + 1)}] - \frac{4}{x^{4} + 1} \\ = \frac{4}{x^{4} - 1} - \frac{4}{x^{4} + 1} \\ = \frac{4 (x^{4} + 1) - 4 (x^{4} - 1)}{(x^{4} - 1) (x^{4} + 1)} \\ = \frac{8}{x^{8} - 1} \end{aligned}

$\eqalign{ & {\text{Given expression,}} \cr & \left( {\frac{1}{{x - 1}} - \frac{1}{{x + 1}}} \right) - \frac{2}{{{x^2} + 1}} - \frac{4}{{{x^4} + 1}} \cr & = \left[ {\frac{{\left( {x + 1} \right) - \left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {x + 1} \right)}}} \right] - \frac{2}{{{x^2} + 1}} - \frac{4}{{{x^4} + 1}} \cr & = \left( {\frac{2}{{{x^2} - 1}} - \frac{2}{{{x^2} + 1}}} \right) - \frac{4}{{{x^4} + 1}} \cr & = \left[ {\frac{{2\left( {{x^2} + 1} \right) - \left( {{x^2} - 1} \right)}}{{\left( {{x^2} - 1} \right)\left( {{x^2} + 1} \right)}}} \right] - \frac{4}{{{x^4} + 1}} \cr & = \frac{4}{{{x^4} - 1}} - \frac{4}{{{x^4} + 1}} \cr & = \frac{{4\left( {{x^4} + 1} \right) - 4\left( {{x^4} - 1} \right)}}{{\left( {{x^4} - 1} \right)\left( {{x^4} + 1} \right)}} \cr & = \frac{8}{{{x^8} - 1}} \cr}$