Let a field $\mathbb{K}$ and $a \in \mathbb{K}^*$ such that $a^{12} = e$. I have to show that $(a+a^{-1})^2 = 3$
Since $\mathbb{K}$ is a field so it is commutative and by binomial theorem and $aa^{-1} = e$
$$ (a + a^{-1})^2 = a^2 + 2e + (a^{-1})^2 $$
To get $(a + a^{-1})^2 =3 $ we have to show $a^2 + (a^{-1})^2 = e$ i .e. I have to find $ x \in \mathbb{K}^* $, $a^2 + (a^{-1})^2 = xx^{-1} = e$ To find such $x$, we have $a^{12} = e$ so $a^{-1} = a^{11}$ but I get $$ a^2 + (a^{-1})^2 = a^2+a^{22} = a^2 + a^{12}a^{10} = a^2 + a^{10} $$ And I don't know how to get further..
Thank you for your help.