Rational number solution for an equation Does there exist $v=(a,b,c)\in\mathbb{Q^3}$ with none of $v$'s terms being zero s.t. $  a+b\sqrt[3]2+c\sqrt[3]4=0$ ?
And I was doing undergraduate algebra 2 homework when I encountered it in my head. At first It seemed like it can be proved there can be no such $v$ like how $\sqrt{2}$, or $\sqrt{2}+\sqrt{3}$ are proved to be irrational, but this case wasn't easy like those. Or maybe I was too hasty.
 A: I'll try to show some more elementary approach, not using field theory. Suppose there is such a $v$. We have $c \ne 0$ (as $\sqrt[3]2$ is irrational). Rewriting the equation, we find $\alpha, \beta \in \mathbb Q$ with 
\[ \sqrt[3]4 = \alpha + \beta \sqrt[3]2 \]
and $\alpha, \beta \ne 0$ (as $\sqrt[3]2, \sqrt[3]4 \not\in \mathbb Q$). 
Taking the third power, we get 
\[ 4 = \alpha^3 + 3\alpha^2\beta \sqrt[3]2 + 3\alpha\beta^2\sqrt[3]4 + 2\beta^3 \]
so, as $\alpha\beta^2 \ne 0$, 
\[ \sqrt[3]4 = \frac{4 - \alpha^3 - 3\alpha^2\beta\sqrt[3]2 - 2\beta^3}{3\alpha\beta^2} \]
Which gives 
\[ \alpha + \beta \sqrt[3]2 = \frac{4-\alpha^3 - 2\beta^3}{3\alpha\beta^2} - \frac\alpha\beta \sqrt[3]2 \]
As $\sqrt[3]2$ is irrational, we must have
\[ \alpha = \frac{4-\alpha^3 - 2\beta^3}{3\alpha\beta^2}, \beta = -\frac\alpha\beta \]
So $\beta^2 = -\alpha$, giving
\[ -3\alpha^3 = 3\alpha^2\beta^2 = 4-\alpha^3 - 2\beta^3 \iff 2(\beta^3 - \alpha^3) = 4 \iff \beta^3 - \alpha^3 = 2 \iff \beta^3 + \beta^6 = 2
\]
But $x^6 + x^3 - 2$ has no rational zeros, as $\pm 1, \pm 2$ are the only possibilities. Contradiction.
So, there is no such $v$.
A: Suppose there exists such a v.
Then $a = - b\sqrt[3]2 - c\sqrt[3]4 $
You can try to show that the left hand side is irrational. Which would show there is no such v.
If there really is no such v.
A: The question can be rephrased as:

Is the triple  $\ 1,\sqrt[3]2,\sqrt[3]4\ $  linearly independent over $\mathbb Q$?

And the answer is  no (i.e., there is no such $v$).
Denote $\alpha:=\sqrt[3]2$. Then $\alpha^3=2$. The main point is, that the polynomial $x^3-2$ (which defines $\alpha$ as its root) is irreducible over $\mathbb Q$: cannot be written as a proper product of polynomials of degree 1 and 2. (This can be shown directly..)
In other words, the field extension $\mathbb Q(\alpha)$ of $\mathbb Q$, is --as $\alpha$ is the root of the irreducible $x^3-2$-- per definition, is isomorphic to the quotient $K:=\mathbb Q[x]/(x^3-2)$ of polynomial ring $\mathbb Q[x]$. That is, the elements of $K$ are the polynomials, but $x^3-2 = 0$ is assumed (as the only rule) in $K$.
And, similarly as $\mathbb Q[x]$ has $1,x,x^2,x^3,x^3,\ldots$ as basis,  $K$ has $1,x,x^2$ as a (standard) basis over $\mathbb Q$. ($x^3$ and above powers can be rephrased by $1,x,x^2$, using the rule: for example $x^3=2,\ x^4=2x,$ etc.)
The correspondence between $K$ and $\mathbb Q(\alpha)$ is simply given by $x\mapsto\alpha$.
