Does there exist rational $a,b,c$, such that $\sqrt[3]{1}+\sqrt[3]{2}+\sqrt[3]{4}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}$ 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?
 A: This isn't a complete answer but I'm curious to see if this strategy could be made to work:

Suppose there existed such a triple. Note that 
$$w = \sqrt[3]{1}+\sqrt[3]{2}+\sqrt[3]{4} \in \mathbb{Q}\left[\sqrt[3]{2}\right]=K.$$
Since $a$, $b$, and $c$ are rational, we can pull out cubes and write 
$$\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=q_{a}\sqrt[3]{a'}+q_{b}\sqrt[3]{b'}+q_{c}\sqrt[3]{c'}$$
where $a'$, $b'$, $c'$ are cube-free integers and $q_{a}$, $q_{b}$ $q_{c}$ $\in \mathbb{Q}$. Thus, $w \in \mathbb{Q}\left[\sqrt[3]{a'},\sqrt[3]{b'},\sqrt[3]{c'}\right]=L$. Since $\mathbb{Q}[w]=\mathbb{Q}[\sqrt[3]{1}+\sqrt[3]{2}+\sqrt[3]{4}]=\mathbb{Q}[\sqrt[3]{2}]=K$, we see that $L$ extends $K$.

$\Delta_{K}=-108=-2^2\cdot3^3$. I wrote some code to compute $\Delta_{L}$ for various $a'$, $b'$, $c'$ and noticed that the primes dividing $a'$, $b'$, $c'$ always appear in factorization of $\Delta_{L}$ along with an additional factor of at least $3^3$. For instance, if $(a',b',c')=(3,5,7)$ then $\Delta_{L}=-3^{31}\cdot 5^{18} \cdot 7^{18} \cdot 11^{18}$.
A: Denote $t=\sqrt[3]2,w=e^{\frac{2\pi i}{3}}=-\frac{1}{2}+\frac{\sqrt3}{2}i,A=\sqrt[3]a,B=\sqrt[3]b,C=\sqrt[3]c.$
If $A,B,C$ are all $\notin \mathbb{Q},$ rewrite the equation as 
$$A+B+C-t-t^2=1.\tag1$$

Lemma: Every number $g(\theta)$ of the field $K(\theta)$ is likewise an algebraic number over $k$ of degree at most $n$. The relative conjugates of a number $a=g(\theta)$ are the distinct numbers among the numbers $g(\theta_i)\ (i=1,2,\dots, n)$. Each conjugate to $a$ appears equally often among the $g(\theta_i).$ 

You can find this lemma in Erich Hecke, Lectures on the theory of algebraic numbers, page 61.
Denote $K=\mathbb{Q}(A,B,C,t)=\mathbb{Q}(\theta)$ for some algebraic number $\theta.$ Assume $[K:\mathbb{Q}]=n=3^r,$ where $r$ is an integer and $1\leq r \leq4.$
Now the conjugates to $A$ with respect to $\mathbb{Q}$ are $A,Aw,Aw^2,$ so do $B,C,t.$ If $A=g_1(\theta),$ and $\theta_i\ (i=1,2,\dots, n)$ are the conjugates to $\theta$ with respect to $\mathbb{Q},$ then $g_1(\theta_i) (i=1,2,\dots, n)$ are the conjugates to $A$ with respect to $\mathbb{Q},$ and each of $A,Aw,Aw^2$ appears equally often among the $g_1(\theta_i),$ namely $3^{r-1}$ times.
Assume that $A+B+C-t-t^2=g_1(\theta)+g_2(\theta)+g_3(\theta)-g_4(\theta)-g_4(\theta)^2=G(\theta).$ If $\theta$ goes over $\theta_i\ (i=1,2,\dots, n)$ then $g_1(\theta)$ goes over $A,Aw,Aw^2,$ and each of them appears $3^{r-1}$ times. So do $B,C,t.$
Since $A,B,C,t$ are all $\notin \mathbb{Q},$we have $$\sum_{i=1}^{n}g_1(\theta_i)=\frac{n}{3}\sum_{i=0}^{2}Aw^i=0.$$
So do $B,C,t.$ Hence we get $$\sum_{i=1}^{n}G(\theta_i)=0.$$
But $$\sum_{i=1}^{n}G(\theta_i)=\sum_{i=1}^{n}1=n,$$
a contradiction. Hence $A,B,C$ are all $\notin \mathbb{Q}$ is impossible. WLOG we assume that $A \in \mathbb{Q}.$ If $B,C$ are all $\notin \mathbb{Q},$ then $$\sum_{i=1}^{n}G(\theta_i)=nA=n, A=1.$$
Now rewrite $(1)$ to $$B+C-t-t^2=0.$$
We get $$B/t+C/t-t/t-t^2/t=0.$$
$$\sqrt[3]{\frac{b}{2}}+\sqrt[3]{\frac{c}{2}}-t=1.$$
For the same reason we get $\dfrac{b}{2}=1$ or $\dfrac{c}{2}=1.$ Now we are done.
This method can easily generalize to other cases. For example, we can prove that $\sqrt[3]{3},\sqrt[5]{23},\sqrt[11]{24}$ are linearly independent in $\mathbb{Q}.$
