Multiply each term on both sides by \(1+{x}^{a-b}\).
\[\frac{1}{1+{x}^{a-b}}(1+{x}^{a-b})+\frac{1}{1+{x}^{b-a}}(1+{x}^{a-b})=1+{x}^{a-b}\]
Cancel \(1+{x}^{a-b}\).
\[1+\frac{1}{1+{x}^{b-a}}(1+{x}^{a-b})=1+{x}^{a-b}\]
Simplify \(\frac{1}{1+{x}^{b-a}}(1+{x}^{a-b})\) to \(\frac{1+{x}^{a-b}}{1+{x}^{b-a}}\).
\[1+\frac{1+{x}^{a-b}}{1+{x}^{b-a}}=1+{x}^{a-b}\]
Cancel \(1\) on both sides.
\[\frac{1+{x}^{a-b}}{1+{x}^{b-a}}={x}^{a-b}\]
Multiply both sides by \(1+{x}^{b-a}\).
\[1+{x}^{a-b}={x}^{a-b}(1+{x}^{b-a})\]
Expand.
\[1+{x}^{a-b}={x}^{a-b}+1\]
Cancel \(1\) on both sides.
\[{x}^{a-b}={x}^{a-b}\]
Since both sides equal, there are infinitely many solutions.
Infinitely Many Solutions
[Infinite Solutions]
