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. 
 A: Cube both sides, then consider whether the difference could be identically $0$.
A: If $t^{1/3}$ were a polynomial, then its degree would be at least one (because it is not constant). This would imply 
$$
\lim_{t\to\infty}\frac{t^{1/3}}t\ne0.
$$
But, precisely, the limit above is indeed zero. So $t^{1/3}$ cannot be a polynomial. 
A: If $t^{1/3}$ is a polynomial, then its derivative $\frac{1}{3}t^{-2/3}$ also is. But it's clearly impossible (for example it would not be continuous at $0$).
A: 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.
A: To begin with, the way the question is posed in a somewhat unclear fashion. From an elementary point of view, one may say that $X^{1/3}$ is not a polynomial because polynomials are different sort of expressions.
Maybe the question is whether the function $X\mapsto f(X)=X^{1/3}$ is polynomial in the sense that there is a polynomial $P(X)$ such that $f(x)=P(x)$. There's a number of ways to prove this false, like the fact--borrowing from Joel Cohen's answer--that polynomial functions admit derivatives continuous everywhere, which is not the case for the function $f(X)$.
In my opinion the correct way to rephrase the question is whether there may be a polynomial $P(X)$ such that
$$
P(X)^3=X.
$$
This is impossible by considering degrees: the degree of the left hand polynomial is a multiple of $3$ whereas the degree of $X$ is $1$, not a multiple of $3$.
Yet, we learn from algebra courses that one may consider polynomials with coefficients in more general algebraic structures (commutative rings). If the coefficient ring is a domain, in particular a field, the very same degree argument applies. But in more exotic situations the degree argument is not sufficient, like the example
$$
(1+2X)^2=1\qquad\text{in $(\Bbb Z/4\Bbb Z)[X]$}
$$
shows. Thus, a different sort of argument is required. But I stop here since I believe that this is actually beyond the scopes of the question.
A: If $t^{1/3}$ is a polynomial, it is non-zero and
$$
t^{1/3} = a_n t^n + \dots + a_0
$$
for $n \ge 0$ with $a_n \neq 0$, but we also have
$$
t = (a_n t^n + \dots + a_0)^3 = a_n^3 t^{3n} + \dots + a_0.
$$
Therefore, $a_n^3 = 0$, so $a_n = 0$, which is contradicting the assumption that $a_n \neq 0$.
This proof is different from the others in the sense that it is from an algebraic point of view. It does not only say that $t^{1/3}$ is not a polynomial function. It also says that there is no polynomial whose cube is $t$, which makes sense as a statement even if $t$ is not a real number or a function. In algebra, $t$ would be called "an indeterminate". Indeterminates are useful mostly when you want to plug in different kinds of things rather than just real numbers. Their main use is in algebraic theories though.
Hope that helps, 
