Trying this for a while and, I found out that if there is another solution $(a_0,b_0,c_0)$, then there is an infinite amount of solutions $(a_0/2,b_0/2,c_0/2), (a_0/4,b_0/4,c_0/4) \ldots$, but I am not sure that this is a contradiction.
Can someone suggest a new insight or help to prove this statement?
Prove that if $a+b\sqrt[3]{2} + c\sqrt[3]{4} = 0$ then $a=b=c=0$, where $a,b,c$ are rational numbers
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jesuisMuhmd
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If you can show that if for each integer solution $(a,b,c)$ of the equation $(a/2, b/2, c/2)$ is another integer solution, then you are done. (And in fact, we can prove this is the case.) Without assuming $(a, b, c)$ are integer, the implication is trivial and does not lead to any contradiction. – Sangchul Lee Aug 04 '22 at 13:35
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Also https://math.stackexchange.com/q/829005/42969, https://math.stackexchange.com/q/2421073/42969 – found with Approach0 – Martin R Aug 04 '22 at 13:37
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Hint: If $a + b\sqrt[3]{2} + c\sqrt[3]{4} = 0$, then $\sqrt[3]{2}$ is a solution to $a + bx + cx^2 = 0$. What do we know about quadratic equations with rational coefficients?
Klaus
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It seems that this has been asked and answered multiple times before. – Martin R Aug 04 '22 at 13:37
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