field theory - Prove that $x^3+2y^3+4z^3equiv6xyz pmod{7} Rightarrow xequiv yequiv zequiv 0 pmod{7}$

[Old] Wants: $x^3+2y^3+4z^3\equiv6xyz \pmod{7} \Rightarrow x\equiv y\equiv z\equiv 0 \pmod{7}$

[Old] My attempt: With modular arithmetic, I can show that if $x\not\equiv0$, then $y\not\equiv0$, and $z\not\equiv0$; if $x\equiv0$, then $x\equiv y \equiv z \equiv 0 \pmod{7}.$

But I don't see how to show that $x^3+2y^3+4z^3-6xyz\not\equiv0\pmod{7}$ when $x,y,z\not\equiv0$, other than plugging in all possible nonzero values of $x,y,z \pmod{7}$ to derive a contradiction. I hope to prove the desired result with least effort.

Update: As @Lubin pointed out in his answer, "the Norm of a nonzero element of the big field is necessarily nonzero in the base field". But I cannot find a reference on this specific statement. Could anyone tell me why is the quoted sentence above true?

Answer

Someone played a dirty trick on you.

The form $x^3-6xyz + 2y^3+4z^3$ is the “norm form” for the extension $\Bbb F_{7^3}\supset\Bbb F_7$. That is, if you take a generator $\zeta=\sqrt[3]2$ of the big field, which is all right, since $X^3-2$ is irreducible over $\Bbb F_7$, then the Norm of $x+y\zeta+z\zeta^2$ is $w=x^3-6xyz+2y^3+4z^3$, when $x,y,z\in\Bbb F_7$. But the Norm of a nonzero element of the big field is necessarily nonzero in the base field. And that does it.

How did I know this? Many hand computations of the Norm, over the years. Pencil and paper, friends, pencil and paper.