Prove a symmetric reciprocal identity in exponents: Evaluate 1/(1 + x^(b−a) + x^(c−a)) + 1/(1 + x^(a−b) + x^(c−b)) + 1/(1 + x^(b−c) + x^(a−c)).

Introduction / Context:
This classic identity appears in problems involving cyclic symmetry and exponents. Setting suitable substitutions reduces the expression to a known form where the three fractions add to 1. The result is independent of the particular values of a, b, c, provided exponents are defined.

Given Data / Assumptions:

• Expression: S = Σ 1/(1 + x^(difference) + x^(other difference)), cycled over (a,b,c).
• Assume x > 0 (so all powers are defined in reals).

Concept / Approach:
Let p = x^(a−b), q = x^(b−c), r = x^(c−a). Then pqr = x^((a−b)+(b−c)+(c−a)) = x^0 = 1. The sum becomes S = 1/(1 + 1/p + r) + 1/(1 + p + 1/q) + 1/(1 + q + 1/r) after suitable rearrangements. Using pqr = 1, each denominator can be put over a common base and the numerators align to produce a telescoping identity summing to 1.

Step-by-Step Solution:

Verification / Alternative check:
Test with concrete values, e.g., x = 2, (a,b,c) = (1,2,3), compute numerically; the sum evaluates to 1, supporting the identity.

Why Other Options Are Wrong:

• x^(a − b − c), 0, 3: These contradict the established symmetric identity for arbitrary a, b, c.

Common Pitfalls:
Failing to use the crucial relation pqr = 1, or mishandling algebra when clearing denominators.

Final Answer: