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.