There does not exist $a,b \in \mathbb{Q}$ such that $a+b\cdot2^{1/3}=2^{2/3}$ I suspect that $\mathbb{Q}[2^{1/3}]=\{x \in \mathbb{R}; ∃a,b \in \mathbb{Q}(x =a+b\cdot2^{1/3})\}$ is not a field. This is because $2^{1/3} \cdot 2^{1/3} = 2^{2/3}$ is not in $\mathbb{Q}[2^{1/3}]$. However, I am having difficulty proving the statement:
There does not exist $a,b \in \mathbb{Q}$ such that $a+b\cdot2^{1/3}=2^{2/3}$
I have not been able to find a contradiction assuming otherwise. 
 A: If $x^2=a+b\,x,$ we have $x^3=a\,x+b\,x^2=a\,x+b\,(a+b\,x)=a\,b+(a+b^2)\,x$. If $x=2^{1/3},$ we'd have $2^{1/3}=\frac{2-a\,b}{a+b^2},$ a rational number. Let $n$ be the smallest positive integer so that both $n\,2^{1/3}$ and $n\,2^{2/3}$ are integers. Then, $m=n(2^{1/3}-1)=n\,2^{1/3}-n$ is an integer $<n$, and $m\,2^{1/3}=n\,2^{2/3}-n\,2^{1/3}$ and $m\,2^{2/3}=2\,n-n\,2^{2/3}$ are integers, contradicting the minimality of $n$.
A: Let $\zeta=\sqrt[3]2$, and assume $\zeta^2=a+b\zeta$ with $a,b\in\mathbb Q$.
Then
$$ 2 = \zeta(a+b\zeta) = a\zeta+b\zeta^2 = a\zeta+b(a+b\zeta) = (a+b^2)\zeta + ba $$
and because $1$ and $\zeta$ are linearly independent over $\mathbb Q$ we must have $a+b^2=0$ and $ab=2$.
But then $a=-b^2$ and $ab=-b^3$. But there is no $b\in\mathbb Q$ such that $-b^3=2$.
A: We can use $$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-xz-yz),$$
which gives $$a^3+2b^3-4+6ab=0.$$
Now, let $a=\frac{m}{n}$ and $b=\frac{k}{n}$, where $m$, $n$ be integers  and $n$ be natural numbers.
Thus, we get
$$m^3+2k^3-4n^3+6mnk=0,$$ 
which is contradiction because we can make here the infinite descent.
My explanation.
We see that $m$ divided by $2$. Thus, there is $m_1\in\mathbb Z$, for which $m=2m_1$ and we obtain
$$4m_1^3+k^3-2n^3+6m_1nk=0.$$
Now, we see that $k$ divided by $2$, which says that there is $k_1\in\mathbb  Z$, for which $k=2k_1$, which gives
$$2m_1^3+4k^3-n^3+6m_1nk_1=0.$$
Now, we see again that $n$ divided by $2$, which says that there is $n_1\in\mathbb N$, for which $n=2n_1$ and we obtain
$$m_1^3+2k_1^3-4n_1^3+6m_1n_1k_1=0.$$ 
Hence, if there are $(m,n,k)$, where $n$ is a natural numbers, for which 
$$m^3+2k^3-4n^3+6mnk=0$$ then there are $(m_1,n_1,k_1)$ for which this equation holds and $n_1<n$.
Since we can repeat this procedure more and more, we get an infinite sequence of natural numbers  $n>n_1>...$, which is impossible.
Done!
A: Square brackets usually indicate a polynomial ring. For instance, $x^2\in \Bbb Q[x]$. The same is true for $2^{1/3}$ in place of $x$. So your definition of $\Bbb Q[2^{1/3}]$ does not correspond to the standard definition of that notation.
That being said, it is true that $2^{2/3}$ is linearly independent of  the set $\{1,2^{1/3}\}$, so in the two-dimensional vector space over $\Bbb Q$ spanned by $1$ and $2^{1/3}$ does not contain $2^{2/3}$.
