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
 A: Since
$$(a+bx+cx^2)^3\equiv (6bc^2+3a^2c+3ab^2)x^2+(6ac^2+6b^2c+3a^2b)x+(4c^3+2b^3+a^3+12abc)\pmod{(x^3-2)}$$
if $(a+b\sqrt[3]{2}+c\sqrt[3]{2})^3$ belongs to $\mathbb{Q}$ then
$$(2bc^2+a^2c+ab^2)=0, \qquad (2ac^2+2b^2c+a^2b)=0,\tag{1}$$
hence:
$$ a^2 b^2 c^2 = (a^2c+ab^2)(ac^2+b^2 c) = a^3 c^3 + 2a^2 b^2 c^2  + ab^4 c,$$
$$ a^3 c^3 + a^2 b^2 c^2 + ab^4 c = 0,$$
$$ ac (a^2 c^2 + a c b^2 + b^4) = 0.$$
Since the discriminant of $y^2+y+1$ is negative, the last equation implies $a=0$ or $c=0$. If we plug this identities back into $(1)$, we get that the only possibilities are that at least two variables among $\{a,b,c\}$ are zero.
A: Step #$1$:


*

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

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

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


*

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

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

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