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?

We argue by contradiction. Obviously, $a\ne0$ and $b\ne0$. Then $x=\sqrt[3]2$ is a root of the square equation with rational coefficients and hence can be represented in the form
$$
 x=A+\sqrt{B} \qquad\text{or}\qquad x=A-\sqrt B
$$
with rational $A$ and $B\ge0$. In fact, $B>0$, otherwise $x$ would be rational. We have
$$
2=x^3=A^3\pm3A^2\sqrt B+3AB\pm B\sqrt B=A^3+3AB\pm(B+3A^2)\sqrt B
$$
Since $B>0$, it follows that $B+3A^2\ne0$ and hence $\sqrt B$ is rational. Thus, so is $x$. Contradiction.
