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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Instead of making a new question you could have appended your previous one. – Aaron Quitta Sep 08 '17 at 04:27
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@AaronQuitta I believe making a new question is considered to be better practice – Mark Schultz-Wu Sep 08 '17 at 04:32
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1This question asks about integers, but if you look at the answers, some of the deal with rationals: Given that $a+b\sqrt[3]{2} +c\sqrt[3]{4} =0$, where $a,b,c$ are integers. Show $a=b=c=0$. Found using Approach0. (There are probably a few more related questions on this site.) – Martin Sleziak Sep 08 '17 at 04:32
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2This one is precisely this question - just in slightly different formulation: Proving that $\sqrt[3] {2} ,\sqrt[3] {4},1$ are linearly independent over rationals. – Martin Sleziak Sep 08 '17 at 04:35
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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.
Angina Seng
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