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From the extension of my previous question For $a,b,c\in \mathbb{Q}$ \begin{align} a + b \sqrt[3]{2} + c\sqrt[3]{4} =0 \quad \Leftrightarrow \quad a=b=c=0 \end{align} again also one direction is easy, but how about other direction?

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As $X^3-2$ is irreducible over the rationals (Eisenstein's criterion) the least degree of polynomial equation $\alpha=\sqrt[3]2$ satisfies over $\Bbb Q$ is $3$. So, $a+b\alpha+c\alpha^2=0$ has no nontrivial rational solution.

Another approach: suppose $a$, $b$, $c$ do exist, not all zero, with $a+b\alpha+c\alpha^2=0$. We can assume they are integers with no common factor. Then $$-a^3=(b\alpha+c\alpha^2)^3=2b^2+6b^2c\alpha +6bc^2\alpha^2+4c^3 =2b^3+4c^3-6abc.$$ This means that $a$ is even. Then $b+c\alpha+(a/2)\alpha^2=0$. Repeating we get $b$ even, $c$ even, contradicting $a$, $b$, $c$ having no common factor.

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