1. **State the problem:** Given that $a + b + c = 0$, prove that $$\frac{1}{1 + x^a + x^{-b}} + \frac{1}{1 + x^b + x^{-c}} + \frac{1}{1 + x^c + x^{-a}} = 1$$ 2. **Rewrite the problem using the condition:** Since $a + b + c = 0$, we have $c = -a - b$. 3. **Define variables for simplification:** Let $$A = x^a, \quad B = x^b, \quad C = x^c = x^{-a - b} = \frac{1}{x^{a+b}} = \frac{1}{AB}$$ 4. **Rewrite each denominator:** - First denominator: $1 + A + x^{-b} = 1 + A + \frac{1}{B}$ - Second denominator: $1 + B + x^{-c} = 1 + B + \frac{1}{C} = 1 + B + AB$ - Third denominator: $1 + C + x^{-a} = 1 + C + \frac{1}{A}$ 5. **Express the sum:** $$S = \frac{1}{1 + A + \frac{1}{B}} + \frac{1}{1 + B + AB} + \frac{1}{1 + C + \frac{1}{A}}$$ 6. **Simplify denominators:** - First denominator: $$1 + A + \frac{1}{B} = \frac{B + AB + 1}{B}$$ - Third denominator: $$1 + C + \frac{1}{A} = 1 + \frac{1}{AB} + \frac{1}{A} = \frac{AB + 1 + B}{AB}$$ 7. **Rewrite each fraction:** $$\frac{1}{1 + A + \frac{1}{B}} = \frac{B}{B + AB + 1}$$ $$\frac{1}{1 + B + AB}$$ (already simplified) $$\frac{1}{1 + C + \frac{1}{A}} = \frac{AB}{AB + 1 + B}$$ 8. **Notice that denominators of first and third fractions are the same:** $$B + AB + 1 = AB + B + 1$$ So, $$S = \frac{B}{AB + B + 1} + \frac{1}{1 + B + AB} + \frac{AB}{AB + B + 1}$$ 9. **Combine first and third fractions:** $$\frac{B}{D} + \frac{AB}{D} = \frac{B + AB}{D} = \frac{B(1 + A)}{D}$$ where $D = AB + B + 1$. 10. **Sum all fractions:** $$S = \frac{B(1 + A)}{D} + \frac{1}{D} = \frac{B(1 + A) + 1}{D}$$ 11. **Simplify numerator:** $$B(1 + A) + 1 = B + AB + 1 = D$$ 12. **Therefore:** $$S = \frac{D}{D} = 1$$ **Final answer:** $$\boxed{1}$$