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

## Concept & Formula This involves stepwise algebraic simplification of rational expressions. Like the previous problem, it relies heavily on creating common denominators using the difference of squares. Formula: $(a - b)(a + b) = a^2 - b^2$ ## Step-by-Step Solution * Solve the expression in pairs from left to right. Take the first two terms: $\frac{1}{x-1} - \frac{1}{x+1}$ * Find the common denominator $(x-1)(x+1) = x^2 - 1$: $\frac{(x+1) - (x-1)}{x^2 - 1} = \frac{2}{x^2 - 1}$ * Substitute this result back into the expression, taking the next term: $\frac{2}{x^2 - 1} - \frac{2}{x^2 + 1}$ * Find the common denominator $(x^2 - 1)(x^2 + 1) = x^4 - 1$: $\frac{2(x^2 + 1) - 2(x^2 - 1)}{x^4 - 1} = \frac{2x^2 + 2 - 2x^2 + 2}{x^4 - 1} = \frac{4}{x^4 - 1}$ * Substitute this result back, taking the final term: $\frac{4}{x^4 - 1} - \frac{4}{x^4 + 1}$ * Find the common denominator $(x^4 - 1)(x^4 + 1) = x^8 - 1$: $\frac{4(x^4 + 1) - 4(x^4 - 1)}{x^8 - 1} = \frac{4x^4 + 4 - 4x^4 + 4}{x^8 - 1} = \frac{8}{x^8 - 1}$ ## Exam Strategy & Shortcut Notice the predictable doubling pattern. The numerator doubles at every step ($1 \rightarrow 2 \rightarrow 4 \rightarrow 8$), and the denominator powers double with a minus sign ($x^2-1 \rightarrow x^4-1 \rightarrow x^8-1$). Knowing this structural behavior, you can directly leap to $\frac{8}{x^8 - 1}$ without performing manual fraction subtractions. ## Common Pitfall A frequent mistake is mishandling the negative signs during numerator expansion (e.g., getting zero in the numerator). Always use parentheses when subtracting binomials in the numerator to ensure the negative distributes correctly across both terms. ## Final Answer **Therefore, the correct answer is $\frac{8}{x^8-1}$.**