How to prove that $n^{\frac{1}{3}}$ is not a polynomial? I'm reading Barbeau's Polynomials, there's an exercise:

How to prove that $n^{\frac{1}{3}}$ is not a polynomial?

I've made this question and with the first answer as an example, I guess I should assume that:
$$n^{\frac{1}{3}}=a_pn^p+a_{p-1}n^{p-1}+\cdots+ a_0n^0$$
And then I should make some kind of operation in both sides, the resultant difference would be the proof. But I have no idea on what operation I should do in order to prove that. 

Assuming $\sqrt[3]{t} = a_n t^n +a_{n-1}t^{n-1}+\cdots +a_0$ and using $\sqrt[3]{8t}=2\sqrt[3]{t}$, we have $$8^na_n t^{n}+8^{n-1}a_{n-1}t^{n-1}+\cdots + a_0 = 2a_n t^n+2a_{n-1}t^{n-1}+\cdots+2a_0.$$  Use uniqueness of coefficients for polynomial functions on $\mathbb R$ to reach a contradiction.
Alternatively, using $\sqrt[3]{t^3}=t$, we have $$t=a_n t^{3n} +a_{n-1}t^{3n-3}+\cdots +a_0,$$ leading to a similar contradiction.
