TITLE: Show determinant of matrix is non-zero
QUESTION [13 upvotes]: I have $a,b,c\in\mathbb{Q}$ not all zero. ($a^2+b^2+c^2\ne 0$), I want to show that the following determinant is then non-zero. I failed to arrive at an appropriate form of the polynomial. Help please.
$$\left|\begin{bmatrix} a & 2c & 2b\\b & a & 2c\\ c & b & a\end{bmatrix}\right| = a^3+2 b^3-6 a b c+4 c^3$$

Second question, what is the easiest way to argue that $\{1,\sqrt[3]{2},(\sqrt[3]{2})^2\}$ is linearly independent in $\mathbb{Q}$?

Motivation:
Prove that $\mathbb{Q}[\sqrt[3]{2}] = \{a+b\sqrt[3]{2}+c(\sqrt[3]{2})^2\;|\;a,b,c\in\mathbb{Q}\}$ forms a field. 
Proof: Since $\mathbb{Q}[\sqrt[3]{2}] \subset \mathbb{R}$, we prove $\mathbb{Q}[\sqrt[3]{2}]$ is a subfield of $(\mathbb{R},+,\cdot)$  
$\forall (a+b\sqrt[3]{2}+c(\sqrt[3]{2})^2)\in \mathbb{Q}[\sqrt[3]{2}]\backslash\{0\}.$ We want to find $(a+b\sqrt[3]{2}+c(\sqrt[3]{2})^2)$ such that  
$(a+b\sqrt[3]{2}+c(\sqrt[3]{2})^2)(d+e\sqrt[3]{2}+f(\sqrt[3]{2})^2) =$
$ (ad+2ec+2bf)+(ae+bd+2cf)\sqrt[3]{2}+(af+cd+be)(\sqrt[3]{2})^2 = 1$  
Since $\{1,\sqrt[3]{2},(\sqrt[3]{2})^2\}$ is linearly independent (?) over $\mathbb{Q}$, we show there is unique solution to:
$$\begin{bmatrix} a & 2c & 2b\\b & a & 2c\\ c & b & a\end{bmatrix} \cdot \left[\begin{array}{l l} d\\e\\f \end{array}\right] = \left[\begin{array}{l l} 1\\0\\0 \end{array}\right] $$
Which is equivalent in showing the determinant is non-zero
$$\left|\begin{bmatrix} a & 2c & 2b\\b & a & 2c\\ c & b & a\end{bmatrix}\right| = a^3+2 b^3-6 a b c+4 c^3=(?)$$  
By subfield test, 1)2)3)4) is enough to say that $(\mathbb{Q}[\sqrt[3]{2}],+,\cdot)$ is a subfield of $(\mathbb{R},+,\cdot)$ therefore a field.
EDIT: If you have shorter way that prove the proposition without touching my 2 questions, that is even better.

REPLY [2 votes]: Just for the linear independence part over $\Bbb Q$. I suppose you know that $\alpha=\sqrt[3]2$, which is by definition the (positive) real root of $X^3-2$, is irrational. But then $X^3-2$ has no rational roots, and (being of degree $3$) is irreducible over $\Bbb Q$, which means it is the minimal polynomial over $\Bbb Q$ of any of its (complex) roots, since such a minimal polynomial has to divide $X^3-2$. In particular the minimal polynomial of $\alpha$ has degree $3$, which implies that $1,\alpha,\alpha^2$ are $\Bbb Q$-linearly independent.