No rational solutions of a system of equations 
Please show that there does not exist $(a,b,c)\in\mathbb{Q}^3$ such that
  \begin{matrix}
a^2b+2b^2c+2ac^2=0\\
a^2c+ab^2+2bc^2=0\\
a^3+2b^3+4c^3+12abc=3.
\end{matrix}

I'm able to show that this is true iff $[\mathbb{Q}(\sqrt[3]2,\sqrt[3]3):\mathbb{Q}]=9$.
 A: In order to prove that there is no $(a,b,c)\in\mathbb{Q}^3$ satisfying the three equations, assume that one of $a,b,c$ is non-zero. Let's say that $a\neq 0$. Set $x=b/a$ and $y=c/a$ and divide the first two equations by $a^3$. Then we get $x+2x^2y+2y^2=0$ and $y+x^2+2xy^2=0$. Multiply the first equation by $x$ and substract from the second gives us that $2x^3y=y$ $\Leftrightarrow$ $y(2x^3-1)=0$. Since $x\in\mathbb Q$ we can not have $2x^3-1=0$, so $y=0$. It follows that $c=0$, and so on.
A: Just to add about why $|\mathbb{Q}(\sqrt[3]{2},\sqrt[3]{3}):\mathbb{Q}|=9$, there are at least a couple of ways to go about approaching it.
Showing that a basis consists of $9$ elements.
You can make an intuitive guess about what a possible basis would be. Starting with $1,\sqrt[3]{2},\sqrt[3]{3},\sqrt[3]{2}^2,\sqrt[3]{3}^2$, a guess for the other $4$ basis elements would be $\sqrt[3]{2}\sqrt[3]{3}=\sqrt[3]{6}$, $\sqrt[3]{2}\sqrt[3]{3}^2$, $\sqrt[3]{2}^2\sqrt[3]{3}$, and $\sqrt[3]{2}^2\sqrt[3]{3}^2=\sqrt[3]{6}^2$. It's then just a case of showing that these are linearly independent and spanning.
Showing that $\mathbb{Q}(\sqrt[3]{2},\sqrt[3]{3})=\mathbb{Q}(\alpha)$, where the minimal polynomial of $\alpha$ has degree $9$.
You might be familiar with the fact that $\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2}+\sqrt{3})$. In this case, it also happens that $\mathbb{Q}(\sqrt[3]{2},\sqrt[3]{3})=\mathbb{Q}(\sqrt[3]{2}+\sqrt[3]{3})$, with minimal polynomial $x^9-15x^6-87x^3-125$.
