Prove $2^{1/3} + 2^{2/3}$ is irrational What's the nice 'trick' to showing that the following expression is irrational?
$2^{1/3} + 2^{2/3}$
 A: Here's a slightly more 'sledgehammer' approach: since $x^3-2$ is irreducible over the rationals, the minimal polynomial of its root $z=2^{1/3}$ must be of degree three.  But if $2^{1/3}+2^{2/3}$ were rational, say $\frac ab$, then that would imply that $z+z^2=\frac ab$, or $bz^2+bz-a=0$, contradicting the minimal-degree statement above.
OTOH, we get a nice prize for all this 'heavy machinery': the exact same argument shows in one fell swoop that $a2^{1/3}+b2^{2/3}$ is irrational for all (nonzero) rational $a,b$; in other words, $2^{1/3}$ and $2^{2/3}$ are linearly independent over $\mathbb{Q}$.
A: $2-1=(2^{1/3}-1)(2^{2/3}+2^{1/3}+1)$ Now if $2^{1/3}-1$ is irrational then so is $2^{2/3}+2^{1/3}+1$ because a product of a rational and irrational is irratonal, which then implies that $2^{2/3}+2^{1/3}$ is irrational.If you can prove $2^{1/3}-1$ is irrational than that's it
A: Let $x=\sqrt[3]{2}+\sqrt[3]{4}$ then raising to the $3^{rd}$ power and expanding the binomial on the right:
$$x^3 = (\sqrt[3]{2})^3 + 3 \cdot \sqrt[3]{2} \cdot \sqrt[3]{4} \cdot (\sqrt[3]{2}+\sqrt[3]{4}) + (\sqrt[3]{4})^3 = 6 x + 6$$
By the rational root theorem, the equation $x^3-6x-6=0$ can only have divisors of $6$ as rational roots, but it's easily verified that none of the divisors is in fact a root. Therefore $x$ is irrational.
