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

Solution(By Examveda Team)

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