Let $F \subset K$ be finite fields with $[K:F] = 3$. Show that if $\alpha \in F$ is not a square in $F$, it is not a square in $K$.
My attempt:
$[K:F] = 3 \implies \forall x \in K$, $x = \beta_1a_1 + \beta_2a_2+\beta_3a_3$, where $\beta_1, \beta_2, \beta_3 \in F$ are basis for K. So if $\alpha \in F$ not a square then $x^2-\alpha = 0$ has no solution in F. So then we have to show that $(\beta_1a_1 + \beta_2a_2+\beta_3a_3)^2 - \alpha = 0$ also has no solutions.
I am not sure if I am even headed in the correct direction. Any help is appreciated.
Assume for contradiction that $\alpha$ is a non-square in $F$, but there is a $\beta\in K$ such that $\beta^2 = \alpha$. Then because $\beta$ is a root of $X^2 - \alpha$, we have $[F(\beta):F] = 2$.
But $[K:F] = [K:F(\beta)]\cdot [F(\beta):F]$, which is impossible.