The value of $\frac{1 - x^4}{1 + x} \div \frac{1 + x^2}{x} \times \frac{1}{x(1 - x)}$ is

### Concept & Formula This problem involves the simplification of rational algebraic expressions. Key techniques include factoring the difference of squares ($a^2 - b^2$) and converting division by a fraction into multiplication by its reciprocal. ### Step-by-Step Solution * **Given Expression:** $$\frac{1 - x^4}{1 + x} \div \frac{1 + x^2}{x} \times \frac{1}{x(1 - x)}$$ * **Step 1: Factorize terms** Use the difference of squares formula $a^2 - b^2 = (a-b)(a+b)$ on $1 - x^4$: $$1 - x^4 = (1 - x^2)(1 + x^2)$$ Factor $(1 - x^2)$ again: $$1 - x^4 = (1 - x)(1 + x)(1 + x^2)$$ * **Step 2: Convert division to multiplication** $$\div \frac{1 + x^2}{x} \text{ becomes } \times \frac{x}{1 + x^2}$$ * **Step 3: Combine and simplify** Substitute everything back into the main expression: $$\frac{(1 - x)(1 + x)(1 + x^2)}{1 + x} \times \frac{x}{1 + x^2} \times \frac{1}{x(1 - x)}$$ Cancel $(1 + x)$ from the first fraction: $$= (1 - x)(1 + x^2) \times \frac{x}{1 + x^2} \times \frac{1}{x(1 - x)}$$ Cancel $(1 + x^2)$: $$= (1 - x) \times x \times \frac{1}{x(1 - x)}$$ Cancel $x$ and $(1 - x)$: $$= 1$$ ### Exam Strategy & Shortcut Use the value substitution method to bypass algebra completely. Pick a simple, permissible value for $x$, like $x = 2$ (avoid $x=0, 1, -1$ to prevent zero denominators). Substitute $x=2$: * Term 1: $\frac{1 - 16}{1 + 2} = \frac{-15}{3} = -5$ * Term 2: $\frac{1 + 4}{2} = \frac{5}{2}$. Division by this is $\times \frac{2}{5}$. * Term 3: $\frac{1}{2(1-2)} = \frac{1}{-2}$. Combine: $-5 \times \frac{2}{5} \times \frac{1}{-2} = -2 \times \frac{1}{-2} = 1$. Since it equals a constant 1, it matches Option (a). ### Common Pitfall Performing the multiplication before the division due to misinterpreting BODMAS/PEMDAS rules. Multiplication and division have the exact same precedence and must be evaluated strictly left-to-right unless grouped by parentheses. ### Final Answer **Therefore, the correct answer is 1.**