Proving that $\sqrt[3] {2} ,\sqrt[3] {4},1$ are linearly independent over rationals I was trying to prove that $\sqrt[3] {2} ,\sqrt[3] {4}$ and  $1$ are linearly independent using elementary knowledge of rational numbers. I also saw this which was in a way close to the question I was thinking about. But I could not come up with any proof using simple arguments. So if someone could give a simple proof, it would be great. 
My try:
$a \sqrt[3] {2}+b\sqrt[3] {4}+c=0$ Then taking $c$ to the other side cubing on both sides we get $2a^3+4b^3+6ab(a+b)=-c^3$. I could not proceed further from here.
Apart from the above question i was also wondering how one would prove that $\sqrt{2},\sqrt{3},\sqrt{5},\sqrt{7},\sqrt{11},\sqrt{13}$ are linearly independent. Here assuming $a\sqrt{2}+b\sqrt{3}+c\sqrt{5}+...=0$ and solving seems to get complicated. So how does one solve problems of this type?

First show that $1$ and $\sqrt[3]2$ are linearly independent. (This should be relatively easy.)
Then in order for $1$, $\sqrt[3]2$ and $\sqrt[3]4$ to be linearly dependent we must have
$$\sqrt[3]{4}=a+b\sqrt[3]2$$
for some $a,b\in\mathbb Q$.
(Since $\sqrt[3]{4}$ is a linear combination of $1$ and $\sqrt[3]2$.)
If we multiply the above equation by $\sqrt[3]2$, we get
\begin{gather*}
8=a\sqrt[3]2+b\sqrt[3]4=a\sqrt[3]2+b(a+b\sqrt[3]2)=(a+b^2)\sqrt[3]2+ab\\
8-ab=(a+b^2)\sqrt[3]2
\end{gather*}
Since $1$ and $\sqrt[3]2$ are linearly independent, we get $2-ab=a+b^2=0$.
So we get
\begin{align*}
ab&=8\\
b^2&=-a
\end{align*}
which yields $b^2=-\frac2b$ and $b^3=-8$. The only rational solution is $b=-2$. This would mean that $a=-b^2=-4$. So in the equation $\sqrt[3]{4}=a+b\sqrt[3]2$ we get that the LHS is positive and the RHS is negative, a contradiction.
