If $abc = 1$, then $$ \left(\frac{1}{1 + a + b^{-1}} + \frac{1}{1 + b + c^{-1}} + \frac{1}{1 + c + a^{-1}}\right) = x $$

### Concept & Strategy When evaluating symmetric cyclic fractions given a product constraint (like $abc = 1$), the most rigorous algebraic method is to substitute variables to make the denominators identical. However, because the equation holds true for *any* values of $a, b, c$ that satisfy the constraint, assuming values is the fastest approach. ### Step-by-Step Solution * **Step 1:** Use the given constraint $abc = 1$ to express $c$ and $c^{-1}$ in terms of $a$ and $b$. $c = \frac{1}{ab}$ $c^{-1} = ab$ * **Step 2:** Substitute these into the second and third fractions to create a common denominator. The first fraction remains: $\frac{1}{1 + a + b^{-1}} = \frac{b}{b + ab + 1}$ The second fraction: $\frac{1}{1 + b + c^{-1}} = \frac{1}{1 + b + ab}$ The third fraction: $\frac{1}{1 + c + a^{-1}} = \frac{1}{1 + \frac{1}{ab} + \frac{1}{a}}$ Multiply the numerator and denominator of the third fraction by $ab$: $\frac{ab}{ab(1) + ab(\frac{1}{ab}) + ab(\frac{1}{a})} = \frac{ab}{ab + 1 + b}$ * **Step 3:** Add the newly formatted fractions together. Notice they now all share the exact same denominator: $(1 + b + ab)$. Sum = $\frac{b}{1 + b + ab} + \frac{1}{1 + b + ab} + \frac{ab}{1 + b + ab}$ Sum = $\frac{1 + b + ab}{1 + b + ab}$ * **Step 4:** Simplify the final fraction. $\frac{1 + b + ab}{1 + b + ab} = 1$ ### Exam Strategy & Shortcut **Value Assumption:** This is a classic cyclic symmetry problem. Since $abc = 1$, simply choose the easiest numbers that fit the rule. Let $a = 1, b = 1, c = 1$. Substitute these directly into the expression: $\frac{1}{1 + 1 + 1^{-1}} + \frac{1}{1 + 1 + 1^{-1}} + \frac{1}{1 + 1 + 1^{-1}}$ $ = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} $ $ = \frac{3}{3} = 1 $ This takes fewer than 10 seconds and guarantees the correct result. ### Common Pitfall Students attempting the algebraic method often get lost in complex nested fractions (like $1 / (1 + 1/ab + 1/a)$) and make simple arithmetic errors when multiplying terms to clear the denominators. Always verify with the substitution method if time permits. ### Final Answer Therefore, the correct answer is **1**.