If rational $a,b,c$ are such that $a\sqrt[3]{3}+b\sqrt[3]{4}+c\sqrt[3]{5}=0$, then $a=b=c=0$ Problem:

Let $a,b,c \in\mathbb{Q}$ such that $a \sqrt[3]{3}+b \sqrt[3]{4}+c\sqrt[3]{5} = 0$.
  Show that $a=b=c=0$.

Case if $a,b,c \in\mathbb{Z}$ is easy, I solved, but I don't have any idea how can I solve the case when $a,b,c \in\mathbb{Q}$. Maybe with polynomial $f = a \sqrt[3]{3}X^3+b \sqrt[3]{4}X^2+c \sqrt[3]{5}X$ and see $f(1) = 0$, but idk.
Any ideas?

Assume one of these three is not $0$. Let's say $c$.
So $a=0, b=0,$ and $c= \frac{p}{q},$ where $p,q \in \mathbb{Z}$
Our equation $\implies \frac{p}{q}\sqrt[3]{5}=0 \implies \frac{p}{q} = 0$
Since $q \neq 0 \implies p=0$
But then $c=\frac{0}{q} =0.$
(Is there something wrong with what I did?)
