Please correct any mistakes in this proof and, if you're feeling inclined, please provide a better one where "better" is defined by whatever criteria you prefer. 


Assume $2^{1/2}$ is irrational. 
$2^{1/3} * 2^{x} = 2^{1/2} \Rightarrow x = 1/6$. 
$2^{1/3} * {2^{1/2}}^{1/3} = 2^{1/2}$.
if $2^{1/2}$ is irrational, then ${2^{1/2}}^{1/3}$ is irrational. 
$2^{1/3} = 2^{1/2} / {2^{1/2}}^{1/3}$. 
$2^{1/3}$ equals an irrational number divided by an irrational number. 
$2^{1/3}$ is an irrational number.


Just use the rational root test on the polynomial equation $x^3-2=0$ (note that $\sqrt[3]{2}$ is a solution to this equation).  If this equation were to have a rational root $\frac{a}{b}$ (with $a,b\in \mathbb{Z}$ and $b\not=0$), then $b\vert 1$ and $a\vert 2$.  Thus, $\frac{a}{b}\in\{\pm 1,\pm 2\}$.  However, none of $\pm 1,\pm 2$ are solutions of $x^3-2=0$.  Therefore the equation $x^3-2=0$ has no rational solutions and $\sqrt[3]{2}$ is irrational.

Alternatively, suppose we have $\sqrt[3]{2}=\frac{a}{b}$ for some $a,b\in \mathbb{Z}$, $b\not=0$, and $\gcd(a,b)=1$.  Then, rearranging and cubing, we have $2b^3=a^3$.  Therefore $a^3$ is even....what does that say about $a$?  What, in turn, does that say about $b$?  It's really not that different from the classic proof that $\sqrt{2}$ is irrational.
