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

I can't resist: Suppose $2^{\frac{1}{3}}=\frac{n}{m}$. Then $$2m^3=n^3,$$ or in other words $$m^3+m^3=n^3.$$ But this contradicts Fermats Last Theorem.

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.

Suppose $2^{1/3}$ is rational. Then $2 \cdot m^3 = n^3$ for some $m, n \in \mathbb{N}$ . Looking at the left side, the power of two in the prime factorization of $2 \cdot m^3$ is of the form $3k + 1$. On the right side, it must be of the form $3l$. This is a contradiction, because the factorizations on both sides must be the same by the fundamental theorem of arithmetic. Thus $2^{1/3}$ cannot be rational.