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?

Consider $c_1\sqrt{2}+c_2\sqrt{3}+c_3\sqrt{5}=0$. Then $c_1\sqrt{2}+c_2\sqrt{3}=-c_3\sqrt{5}$. Squaring both sides we will have $2c_1^2+3c_2^2+2\sqrt{6}c_1c_2=5c_3^2$. If either $c_1$ or $c_2$ turns out to be $0$ then we will either have $c_2\sqrt{3}+c_3\sqrt{5}=0$ implying $3c_2^2=5c_3^2$ which gives $\left(\frac{c_2}{c_3}\right)^2=\frac{5}{3}$ which is not possible . Similarly for the case when $c_2$ is $0$. (It is obvious when both $c_1$ and $c_2$ are $0$.) Hence if $c_1$ and $c_2$ are both  non-zero then $$-\sqrt{6}=\frac{2c_1^2+3c_2^2-5c_3^2}{2c_1c_2}.$$
Now observe that the R.H.S is a rational no but the L.H.S is not. 
