Why is $r^3+4t^3+2s^3-6rts$ non-zero (unless $r=s=t=0$)? When solving How to find the multiplicative inverse of a polynomial? I created a $3 \times 3$ linear system with determinant $r^3+4t^3+2s^3-6rts, \text{ where } r, t, s \in \Bbb Q$.  Basic field theory tells me this determinant must be non-zero (unless $r=s=t=0$), but there must be a more direct way to see that.  Can anyone explain why this quantity must be non-zero in all but the trivial case?  Thanks.
 A: This determinant is the norm of the element $\alpha = r + s \sqrt[3]{2} + t \sqrt[3]{4} \in \mathbb{Q}(\sqrt[3]{2})$. It must be nonzero because $\mathbb{Q}(\sqrt[3]{2})$ is a field, as you say; somewhat more explicitly, it's the product of the conjugates
$$(r + s \sqrt[3]{2} + t \sqrt[3]{4})(r + s \sqrt[3]{2} \omega + t \sqrt[3]{4} \omega^2)(r + s \sqrt[3]{2} \omega^2 + t \sqrt[3]{4} \omega)$$
of $\alpha$, where $\omega = e^{ \frac{2 \pi i}{3} }$ is a primitive third root of unity. This product is nonzero because each of its factors is nonzero because $\mathbb{Q}(\sqrt[3]{2})$ is a field and so are its Galois conjugates $\mathbb{Q}(\sqrt[3]{2} \omega), \mathbb{Q}(\sqrt[3]{2} \omega^2)$.
The three factors above are the three eigenvalues of the matrix of $\alpha$ acting by left multiplication on $\mathbb{Q}(\sqrt[3]{2})$, regarded as a $3$-dimensional vector space over $\mathbb{Q}$ with basis $\{ 1, \sqrt[3]{2}, \sqrt[3]{4} \}$. The determinant of this matrix is (by definition) the norm $N(\alpha)$. Inverting this matrix in order to invert $\alpha$ is, I imagine, where your linear system comes from.
Homogeneous polynomials arising from norms in this way are called norm forms.
A: Let $x = r$, $y =\sqrt[3]{4}t$ and $z = \sqrt[3]{2}s$
$$r^3 + 4t^3 +2s^3 - 6rts = \left(r + t\sqrt[3]{4} + s\sqrt[3]{2}\right)\left(x^2 + y^2 + z^2 - xy -yz - zx\right)$$
$r + t\sqrt[3]{4} + s\sqrt[3]{2} = 0 \iff r = s = t = 0$
$$x^2 + y^2 + z^2 - xy - yz - zx = \frac{1}{2}(x-y)^2 + \frac{1}{2}(y-z)^2 + \frac{1}{2}(z-x)^2$$
so $x^2 + y^2 + z^2 - xy - yz - zx = 0 \iff r = s = t = 0$
