Discussion

Explanation:

Given, $\frac{1}{1+\mathrm{a}+{\mathrm{a}}^2}$ + $\frac{1}{{\mathrm{a}}^{-1}+1+\mathrm{a}}$ + $\frac{1}{{\mathrm{a}}^{-2}+{\mathrm{a}}^{-1}+1}$.

= $\frac{1}{1+\mathrm{a}+{\mathrm{a}}^2}$ + $\frac{1}{{\displaystyle \frac{1}{a}}+1+\mathrm{a}}$ + $\frac{1}{{\displaystyle \frac{1}{a^2}}+{\displaystyle \frac{1}{a}}+1}$.

= $\frac{1}{1+\mathrm{a}+{\mathrm{a}}^2}$ + $\frac{\mathrm{a}}{1+\mathrm{a}+{\mathrm{a}}^2}$ + $\frac{{\mathrm{a}}^2}{1+\mathrm{a}+{\mathrm{a}}^2}$

= $\frac{1+\mathrm{a}+{\mathrm{a}}^2}{1+\mathrm{a}+{\mathrm{a}}^2}$

= 1

Hence, option (c).

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