Rational number trivial case Let $a,b,c$ denote rational numbers, such that $(a+b\sqrt[3]2+c\sqrt[3]4)^3$ is also rational. Prove that at least two of the numbers $a,b,c$ must be zero. Actually I confused of the beginning steps for this proof. Should I apply the strategies of contradictory proofs

Step #$1$:


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*$a+b\sqrt[3]{2}\neq0 \implies c\notin\mathbb{Q}$

*$a+c\sqrt[3]{4}\neq0 \implies b\notin\mathbb{Q}$

*$b\sqrt[3]{2}+c\sqrt[3]{4}\neq0 \implies a\notin\mathbb{Q}$
Step #$2$:


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*$a+b\sqrt[3]{2}=0 \wedge a\neq0 \implies b\notin\mathbb{Q}$

*$a+c\sqrt[3]{4}=0 \wedge a\neq0 \implies c\notin\mathbb{Q}$

*$b\sqrt[3]{2}+c\sqrt[3]{4}=0 \wedge b\neq0 \implies c\notin\mathbb{Q}$
