$$ \frac { 1 } { 1 + x ^ { a } + x ^ { - b } } + \frac { 1 } { 1 + x ^ { b } + x ^ { - c } } + \frac { 1 } { 1 + x ^ { c } + x ^ { - a } } = 1 $$

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The equation you've provided is: $$ \frac{1}{1 + x^{a} + x^{-b}} + \frac{1}{1 + x^{b} + x^{-c}} + \frac{1}{1 + x^{c} + x^{-a}} = 1 $$ To analyze and solve this equation, let's consider the variables involved: $$ x $$, $$ a $$, $$ b $$, and $$ c $$. Without additional constraints or relationships between $$ a $$, $$ b $$, and $$ c $$, we can explore some general properties and potential solutions. ### **1. Symmetric Solution: $$ x = 1 $$** One obvious solution is $$ x = 1 $$. Let's verify this: - When $$ x = 1 $$: $$ x^{a} = 1^{a} = 1, \quad x^{-b} = 1^{-b} = 1, \quad \text{similarly for other terms.} $$ - Substituting into the equation: $$ \frac{1}{1 + 1 + 1} + \frac{1}{1 + 1 + 1} + \frac{1}{1 + 1 + 1} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1 $$ Thus, $$ x = 1 $$ **always** satisfies the equation, regardless of the values of $$ a $$, $$ b $$, and $$ c $$. ### **2. Uniqueness of the Solution $$ x = 1 $$** To determine if $$ x = 1 $$ is the **only** solution, consider the behavior of the function for $$ x \neq 1 $$: - **For $$ x > 1 $$:** - The terms $$ x^{a} $$ and $$ x^{b} $$ (assuming $$ a, b > 0 $$) grow larger, while $$ x^{-b} $$ and $$ x^{-c} $$ become smaller. - The denominators $$ 1 + x^{a} + x^{-b} $$, $$ 1 + x^{b} + x^{-c} $$, and $$ 1 + x^{c} + x^{-a} $$ increase, causing each fraction to decrease. - The sum of the fractions may become less than 1. - **For $$ 0 < x < 1 $$:** - The terms $$ x^{a} $$ and $$ x^{b} $$ decrease, while $$ x^{-b} $$ and $$ x^{-c} $$ increase. - The behavior of the denominators and hence the fractions will depend on the specific values of $$ a $$, $$ b $$, and $$ c $$. - It's possible for the sum of the fractions to be greater than or less than 1 based on these values. Given these observations, **without specific relationships or constraints between $$ a $$, $$ b $$, and $$ c $$**, it's challenging to assert whether additional solutions exist for $$ x \neq 1 $$. The equation's solution can vary based on the parameters: - **If $$ a $$, $$ b $$, and $$ c $$ satisfy certain symmetry or functional relationships**, other solutions for $$ x $$ might emerge. - **In the absence of such constraints**, $$ x = 1 $$ remains the primary obvious solution. ### **Conclusion** **$$ x = 1 $$ is always a solution to the given equation, regardless of the values of $$ a $$, $$ b $$, and $$ c $$.** Determining whether other solutions exist would require additional information or constraints on the variables $$ a $$, $$ b $$, and $$ c $$. --- **Final Answer:** For all exponents a, b, c, the equation is satisfied when x = 1. Thus, x = 1 is a solution. $$ x = 1 $$ is always a solution to the equation.

\( \frac { 1 } { 1 + x ^ { a } + x ^ { - b } } + \frac { 1 } { 1 + x ^ { b } + x ^ { - c } } + \frac { 1 } { 1 + x ^ { c } + x ^ { - a } } = 1 \)

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