<s>[INST] 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.
 [/INST] 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 $v$.
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