Q:

Prove that if $p$,$q$ and $\sqrt{2}p+\sqrt[3]{3}q$ are rational numbers then $p=q=0$

I would appreciate if somebody could help me with the following problem: 
Q: Prove that if $p$,$q$ and $\sqrt{2}p+\sqrt[3]{3}q$ are rational numbers then  $p=q=0$

A:

Hint: Put $ a = p\sqrt{2} + q\sqrt[3]{3}\Rightarrow (a-p\sqrt{2})^3 = 3q^3\Rightarrow \sqrt{2} = \cdots =\dfrac{M}{N}$ with $M, N \in \mathbb{Z}$.

A:

If $\sqrt{2}p+\sqrt[3]{3}q$ is rational for some rational $p,q$, then $1,\sqrt{2},\sqrt[3]{3}$ are linearly dependent over $\mathbb Q$.
Since $1,\sqrt{2}$ are linearly independent, we must then have that $\sqrt[3]{3} \in \mathbb Q[\sqrt{2}]$.
But $\mathbb Q[\sqrt[3]{3}]$ has dimension $3$ over $\mathbb Q$ and so cannot be a subspace of $\mathbb Q[\sqrt{2}]$, which has dimension $2$.

