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?

Let $\alpha=\sqrt[3] {2}$ and suppose $a+b\alpha+c\alpha^2=0$ with $a,b,c\in \mathbb Z$, which we may assume coprime.
Then $a\alpha+b\alpha^2+2c=0$ and $a\alpha^2+2b+2c\alpha=0$.
This means that the matrix below is singular
$$
\pmatrix{ a & b & c \\ 2c & a & b \\ 2b & 2c & a}
$$
Its determinant must be zero:
$$
a^3-6 a b c+2 b^3+4 c^3=0
$$
This implies that $a$ is even: $a=2A$. So
$$
4A^3-6 A b c+ b^3+2 c^3=0
$$
This implies that $b$ is even: $b=2B$. So
$$
2A^3-6 A B c+ 4B^3+ c^3=0
$$
This implies that $c$ is even. This contradicts they being coprime.
