<s>[INST] If $a,b,c$ are rational numbers and if $\left(a+b\sqrt[3]{2}+c\sqrt[3]{4}\right)^3$ is also rational then prove that $ab+bc+ca=0$ 
If $a,b,c$ are rational numbers and if $\displaystyle \left(a+b\sqrt[3]{2}+c\sqrt[3]{4}\right)^3$ is also rational then prove that $ab+bc+ca=0$

My attempt
Binomial expansion is not a good idea because there will be $27$ terms so I tried to prove using factorization.
$$\displaystyle \left(a+b\sqrt[3]{2}+c\sqrt[3]{4}\right)^3-\left(c\sqrt[3]{4}\right)^3
\\=\left(a+
b\sqrt[3]{2}\right)\left[\left(a+b\sqrt[3]{2}+c\sqrt[3]{4}\right)^2+\left(a+b\sqrt[3]{2}+c\sqrt[3]{4}\right)\left(c\sqrt[3]{4}\right)+\left(c\sqrt[3]{4}\right)^2\right]$$
This again leads to complicated calculations. Then I tried to equate it to a rational number $r$.
\begin{align*} 
\displaystyle \left(a+b\sqrt[3]{2}+c\sqrt[3]{4}\right)^3&=r\\
\implies a+b\sqrt[3]{2}+c\sqrt[3]{4}&=r^{1/3}\\ 
\implies b\sqrt[3]{2}+c\sqrt[3]{4}&=r^{1/3}-a\\
\implies(b\sqrt[3]{2}+c\sqrt[3]{4})^3&=(r^{1/3}-a)^3\\
\implies2b^3+6\sqrt[3]{2}~b^2c+6\sqrt[3]{4}~bc^2+4c^3&=r-3r^{2/3}a+3r^{1/3}a^2-a^3
\end{align*}
When I got stuck here I wrote the equation $a+b\sqrt[3]{2}+c\sqrt[3]{4}=r^{1/3}$ in $3$ different ways, each time multiplying with $\sqrt[3]{2}$
\begin{align*} \displaystyle a+b\sqrt[3]{2}+c\sqrt[3]{4}&=r^{1/3}\\ 
 a\sqrt[3]{2}+b\sqrt[3]{4}+2c&=\sqrt[3]{2}r^{1/3}\\
 a\sqrt[3]{4}+2b+2\sqrt[3]{2}c&=\sqrt[3]{4}r^{1/3}\end{align*}
I tried adding the above three equations but it wasn't helpful.
Can someone help me in solving the question. Thanks in advance.
Is it possible to generalize the question as $\left(a+b\sqrt[3]{n}+c\sqrt[3]{n^2}\right)^3$ where $n$ is a non-square integer?
 [/INST] By using $$(x+y+z)^3=x^3+y^3+z^3+3xy(x+y)+3xz(x+z)+3yz(y+z)+6xyz$$ we get $a^2b+2b^2c+2ac^2=0$ and $ab^2+a^2c+2bc^2=0$. Now multiply the first equation by $b$ and the second by $a$ and then subtract them. We obtain $c=0$ (and then $a=0$ or $b=0$) or $a=b=0$.
Edit. I forgot to mention that I used the following: if $p+q\sqrt[3]{2}+r\sqrt[3]{4}=0$, with $p,q,r$ rational numbers, then $p=q=r=0$.
The generalization is straightforward.
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