\( \Large a^{2} + 1 => a = a + \frac{1}{a} = 1 \)

On squaring both sides, we get

\( \Large a^{2} + \frac{1}{a^{2}}= -1 \)

On cubing both sids, we gt

\( \Large \left(a^{2} + \frac{1}{a^{2}}\right)^{3} = \left(-1\right)^{3} \)

=> \( \Large a^{6} + \frac{1}{a^{6}} + 3a^{2} \times \frac{1}{a^{2}} \left(a^{2} + \frac{1}{a^{2}}\right) = -1 \)

=> \( \Large a^{6} + \frac{1}{a^{6}} + 3 \times \left(-1\right) = -1 \)

Now, \( \Large a^{6} + \frac{1}{a^{6}} + 1 = 3 \)

As \( \Large a^{12} + a^{6} + 1 \) can also be written as

\( \Large a^{6} + \frac{1}{a^{6}} + 1 \)

Therefore, \( \Large a^{12} + a^{6} + 1 = 3 \)