Question: the problem at hand is:

Prove that there is an irrational number $a$ such that $a{^\sqrt{3}}$ is rational.

Hint: Consider $\sqrt[3]{2}^\sqrt{3}$ and argue by cases.

Any help on this problem would be greatly appreciated!

Answer: 
$\alpha=\sqrt[3]{2}^{\sqrt{3}}$ is trascendental (hence irrational) by the [Gelfond-Schneider theorem](https://en.wikipedia.org/wiki/Gelfond%E2%80%93Schneider_theorem), hence $\alpha^{\sqrt{3}}=2$ proves the statement, but a simple dichotomy is sufficient to answer the question. If $\alpha=\sqrt[3]{2}^{\sqrt{3}}$ is rational we're done, and if $\alpha=\sqrt[3]{2}^{\sqrt{3}}$ is irrational we're equally done by considering $\alpha^{\sqrt{3}}=2$.
Another approach is to compare the cardinalities of the rational numbers and irrational numbers. Assume $f(x)=x^{\sqrt{3}}$ is rational for every irrational number $x$ and define

$$A=\mathbb{R}^{+}-\mathbb{Q}\text{ (positive irrational numbers)}$$

$$B=\{r:f(x)=r\text{ for some }x\in A\}$$

Note that by our assumption $B\subseteq \mathbb{Q}$. Since $f(x)$ is increasing on the positive reals, there is a map from $B$ to $A$. Thus

$$A\subseteq B\subseteq \mathbb{Q}$$

This is a contradiction as $A$ is an uncountable set while the rationals are a countable set. We conclude there exists an irrational number $x$ such that $x^{\sqrt{3}}$ is rational.