$L/K\DeclareMathOperator{\char}{char}$ separable extension of degree $[L:K]=3$. $d(L/K)$ is a discriminant of the any basis of $L/K$. Then $L/K$ is Galois if and only if $d(L/K)$ is a square in $K^\star$.
$\textbf{Q:}$ Do I need $\char(K)\neq 2$ for $d(L/K)$ square implying $L/K$ Galois? In my proof the forward direction does not require this condition. When I try to deduce the opposite direction, I do need $\char(K)\neq 2$. The following is my proof for the opposite direction.
$d(L/K)=(\det(\sigma_i(a^j)))^2$ as $L=K(a)$ for some $a$ by primitive element theorem. Now $\det(\sigma_i(a^j))=(a-a_1)(a-a_2)(a_1-a_2)$ where $a_i$ are conjugates of $a$. Since $g(x)=(x-a_1)(x-a_2)\in K(a)[x]$ by $(x-a)(x-a_1)(x-a_2)\in K(a)[x]$, I have $a_1+a_2\in K(a)$. It suffices to show $a_1-a_2\in K(a)$. Now $g(a)\in K(a)$. Hence $K(a)\ni \frac{\det(\sigma_i(a^j))}{g(a)}=(a_1-a_2)$ where $g(a)\neq 0$ by separability. However I need $\char(K)\neq 2$ to finish this off.
Ref: Algebraic Number Theory by Taylor Frolich Chapter 2 Exercise.1.