Prove $2^{1/3} + 2^{2/3}$ is irrational

What's the nice 'trick' to showing that the following expression is irrational?

$2^{1/3} + 2^{2/3}$

Solution 1:

$2-1=(2^{1/3}-1)(2^{2/3}+2^{1/3}+1)$ 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

Solution 2:

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.