TITLE: Does there exist rational $a,b,c$, such that $\sqrt[3]{1}+\sqrt[3]{2}+\sqrt[3]{4}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}$
QUESTION [36 upvotes]: Let $w = \sqrt[3]{1}+\sqrt[3]{2}+\sqrt[3]{4}$.
How to prove that there are no triples $(a,b,c)$, such that

$a,b,c \in \mathbb{Q}$;
$a \leqslant b \leqslant c$;
$(a,b,c)\ne (1,2,4)$;
$w = \sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}$.

Or maybe there exists one?

REPLY [25 votes]: Everything will follow from the iteration of the following lemma (whose proof is explained
at the very end of this answer) :
 FUNDAMENTAL LEMMA.  Let ${\mathbb K}$ be a field, and let
$a$ be a non-cube in ${\mathbb K}$. Then all the “new” cubes  in ${\mathbb K}(\sqrt[3]a)$ 
are made up of $a$ and “old” cubes : they can be written as $w^3,aw^3$ or $a^2(w^3)$ for some $w\in {\mathbb K}$.
 COROLLARY 1.  Let ${\mathbb K}$ be a subfield of $\mathbb R$, and let $a_1,a_2, \ldots, a_n$ be a sequence of real numbers such that for any $i$, $a_i$ is a non-cube
in ${\mathbb K}(\sqrt[3]{a_1},\sqrt[3]{a_2}, \ldots ,\sqrt[3]{a_{i-1}})$. Then all the cubes in ${\mathbb K}(\sqrt[3]{a_1},\sqrt[3]{a_2}, \ldots ,\sqrt[3]{a_{n}})$ can be written as $a_1^{i_1}a_2^{i_2} \ldots a_n^{i_n}(w^3)$ where $w\in {\mathbb K}$ and each
$i_k$ is $0,1$ or $2$.
 COROLLARY 2.  Let $r_1,r_2, \ldots ,r_n$ be positive rational numbers
 such that there is no-nontrivial relationship between them of the form
 $r_1^{i_1}r_2^{i_2} \ldots r_n^{i_n}=w^3$, where each
$i_k$ is $0,1$ or $2$, and $w$ is rational (non-trivial means that not all the $i_k$ are zero). Then the $3^n$ algebraic numbers $s_1^{i_1}s_2^{i_2} \ldots s_n^{i_n}$ (where
$s_i=\sqrt[3]{r_i}$) are linearly independent over the rationals.
Now, suppose that $\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=\sqrt[3]{1}+\sqrt[3]{2}+\sqrt[3]{4}$. The degree 
of the extension ${\mathbb Q}(\sqrt[3]{2},\sqrt[3]{4},\sqrt[3]{a},\sqrt[3]{b},\sqrt[3]{c})$
(call it $d$) is a power of three. If it were strictly greater than $3$, then $d\geq 9$
and by corollary 2, any non-trivial relation between cube roots would need at least nine
terms, which is too much for the six we've got. So $d=3$, which means that $a,b,c$ must all be, up to sign,  (possibly negative) powers of $2$.  So there are essentially three cases :
$$
\begin{eqnarray}
\sqrt[3]{2^i}+\sqrt[3]{2^j}+\sqrt[3]{2^k}&=& \sqrt[3]{1}+\sqrt[3]{2}+\sqrt[3]{4} \\
\sqrt[3]{2^i}+\sqrt[3]{2^j}-\sqrt[3]{2^k}&=& \sqrt[3]{1}+\sqrt[3]{2}+\sqrt[3]{4} \\
\sqrt[3]{2^i}-\sqrt[3]{2^j}-\sqrt[3]{2^k}&=& \sqrt[3]{1}+\sqrt[3]{2}+\sqrt[3]{4} \\
\end{eqnarray}
$$
where $i,j,k$ are integers. It is easy now to see then that $i,j,k$ must form a complete
system of residues modulo $3$, and to check that $(1,2,4)$ is indeed the only solution
to the original problem.
 PROOF OF FUNDAMENTAL LEMMA . Let ${\mathbb L}={\mathbb K}(\sqrt[3]a)$. We have 
$[{\mathbb L}:{\mathbb K}]=3$. Let $c$ be a cube in $\mathbb L$. Then, there are
$x_1,x_2,x_3 \in K$ such that $c=(x_1+x_2\sqrt[3]{a}+x_3\sqrt[3]{a^2})^3$. Expanding, we find $c=C_0+3C_1\sqrt[3]a+3C_2\sqrt[3]{a^2}$, with 
$$
\begin{eqnarray}
C_0&=&\bigg(x_1^3 + 6ax_1x_2x_3 + (ax_2^3 + a^2x_3^3)\bigg), \\
C_1&=&\bigg(x_1^2x_2 + ax_1x_3^2 + ax_2^2x_3\bigg), \\
C_2&=&\bigg(x_1^2x_3 + x_1x_2^2 + ax_2x_3^2\bigg) 
\end{eqnarray}
$$ 
So $C_1$ and $C_2$ must both be $0$, and hence
$$
ax_2x_3(x_2^3-ax_3^3)=(x_2^2+x_1x_3)C_1-(ax_3^2+x_1x_2)C_2=0
$$
Since $a$ is not a cube in $\mathbb K$, we must have $x_3=0$ or $x_2=0$.
If $x_3=0$, $C_2$ simplifies to $x_1x_2^2$, and if $x_2=0$, $C_2$ simplifies
to $x_1^2x_3$. So at least two of $x_1,x_2,x_3$ are zero, which means that $x_1+x_2\sqrt[3]{a}+x_3\sqrt[3]{a^2}$ is a multiple (by some element of $\mathbb K$) 
of $1,\sqrt[3]{a}$, or $\sqrt[3]{a^2}$, as wished.