Background: Since $x^3\not\equiv2\pmod{7}, \forall x\in\mathbb{Z}$, we can let $K=\mathbb{F_7}[\sqrt[3]{2}]$ so that $K$ is an extension of $\mathbb{F_7}$, which can be think of as a finite-dimensional vector space over $\mathbb{F_7}$ with a basis $(1,\sqrt[3]{2},\sqrt[3]{4})$. Then the norm of an element $\alpha=x+y\sqrt[3]{2}+z\sqrt[3]{4}$ is defined to be $N(\alpha)=x^3+2y^3+4y^3-6xyz$, which is the determinant of the linear transformation $f_\alpha:K \rightarrow K$ defined as left multiplication by $\alpha$ in $K$.
My question is that if I have an element $\beta\in K$ having a unit norm, i.e. $N(\beta)=1$, can I then conclude that $N(\beta\alpha)=N(\beta)N(\alpha)=N(\alpha)$? I.e., is $N$ a homomorphism?
My attempt Observe that $f_{\beta\alpha}=f_\beta\circ f_\alpha$, then $$N(\beta\alpha)=\det(f_{\beta\alpha})=\det(f_\beta)\det(f_\alpha)=N(\beta)N(\alpha).$$
My motivation: Then $x^3+2y^3+4z^3-6xyz=2$ has infinitely many integral solutions since we can just find an $\alpha$ s.t. $N(\alpha)=2$ and some non-identity $\beta$ s.t. $N(\beta)=1 \Rightarrow \forall n\in\mathbb{N}, N(\beta^n\alpha)=2$, i.e. we have infinitely many solutions to the give cubic form.
Yes $N$ is a homomorphism. One easy way to see this is that determinants are multiplicative.
I.e., if $L/K$ is a field extension, and $\alpha,\beta \in L$, and if $\theta\in L$, then $\mu_\theta : L\to L$ denotes the map $\mu_\theta(x) = \theta\cdot x$, then we have $$N(\alpha \beta) = \det \mu_{\alpha\beta} = \det (\mu_\alpha \circ \mu_\beta) = \det \mu_\alpha \cdot \det \mu_\beta = N(\alpha)N(\beta).$$
Edit
You just added your attempt, and it's what I have here. So good work!