Find the value of   $\frac{1}{1+{\mathrm{x}}^{\mathrm{a}-\mathrm{b}}}+\frac{1}{1+{\mathrm{x}}^{\mathrm{b}-\mathrm{a}}}\mathrm{ }$ is equal to

Simplify $\frac{1}{1+{\mathrm{x}}^{\mathrm{a}-\mathrm{b}}}+\frac{1}{1+{\mathrm{x}}^{\mathrm{b}-\mathrm{a}}}\mathrm{ }$ by using the law of indices.
Use the law of indices ${\mathrm{x}}^{\mathrm{m}-\mathrm{n}}=\frac{{\mathrm{x}}^{\mathrm{m}}}{{\mathrm{x}}^{\mathrm{n}}}$
and rewrite the terms,
= $\frac{1}{1+\frac{{\mathrm{x}}^{\mathrm{a}}}{{\mathrm{x}}^{\mathrm{b}}}}+\frac{1}{1+\frac{{\mathrm{x}}^{\mathrm{b}}}{{\mathrm{x}}^{\mathrm{a}}}}$
Take LCM in the denominator of both fractions,
 = $\frac{1}{{\displaystyle \frac{{\mathrm{x}}^{\mathrm{b}}+{\mathrm{x}}^{\mathrm{a}}}{{\mathrm{x}}^{\mathrm{b}}}}}+\frac{1}{{\displaystyle \frac{{\mathrm{x}}^{\mathrm{a}}+{\mathrm{x}}^{\mathrm{b}}}{{\mathrm{x}}^{\mathrm{a}}}}}$
= $\frac{{\mathrm{x}}^{\mathrm{b}}}{{\mathrm{x}}^{\mathrm{b}}+{\mathrm{x}}^{\mathrm{a}}}+\frac{{\mathrm{x}}^{\mathrm{a}}}{{\mathrm{x}}^{\mathrm{a}}+{\mathrm{x}}^{\mathrm{b}}}\mathrm{ }$
Again, take the LCM of both fractions,
= $\frac{{\mathrm{x}}^{\mathrm{b}}+{\mathrm{x}}^{\mathrm{a}}}{{\mathrm{x}}^{\mathrm{b}}+{\mathrm{x}}^{\mathrm{a}}}$
Cancel out the common factors from the numerator and denominator,
 = 1
Thus, the value of $\frac{1}{1+{\mathrm{x}}^{\mathrm{a}-\mathrm{b}}}+\frac{1}{1+{\mathrm{x}}^{\mathrm{b}-\mathrm{a}}}\mathrm{ }$is 1.