Let $L/K$ be a cubic (or, more generally, odd-order) extension of fields of characteristic $0$. To every element $a \in L^\times$ we can associate the quadratic form \begin{align*} q_a : L &\to K \\ x &\to \operatorname{tr}_{L/K}(ax^2) \end{align*} and then take its class in the $2$-Brauer group $\operatorname{Br}(K)[2]$ (using that Severi-Brauer varieties of order $2$ are conics). This class does not change if $a$ is scaled by an element of $K$ or a square in $L$, and so we get a map $$ e : L^\times / ((L^\times)^2 \cdot K^\times) \to \operatorname{Br}(K)[2], $$ Thus $e(a) = 0$ exactly when $\operatorname{tr}_{L/K}(ax^2) = 0$ for some nonzero $x$. Alternatively, we can view $e$ as a map $$ e : H^1(K, E[2]) \to H^2(K, \mu_2) $$ where $E$ is any elliptic curve over $K$ whose $2$-torsion has field of definition $L$.

The map $e$ is not in general a group homomorphism. Numerical evidence suggests that $e$ is a constant plus a quadratic form whose associated bilinear form is the Hilbert symbol: that is, for all $a,b \in L^\times$ of norm $1$, \begin{equation}\tag{1}\label{eq:main} e(ab) - e(a) - e(b) + e(1) = \langle a, b\rangle, \end{equation} where the Hilbert symbol $\langle a, b\rangle$ is the class in the $2$-Brauer group of the conic $$ a x^2 + b y^2 = z^2. $$ When $L/K$ is Galois, I can prove \eqref{eq:main} by diagonalizing $q_a$ after a base change to $L$. But in the non-Galois case, I'm stuck with a nasty relation between the Hilbert symbols over $L$ and a quadratic extension. If it helps, I'm principally interested in the case when $K \supseteq \mathbb{Q}_2$ is a local field (and hence $\operatorname{Br}(K)[2]$ has just two elements).