How to show that $\sqrt[n]{3}$ is not a real root of a polynomial of degree $<n$, where $n\geq 2$.
Suppose for some $f(x)=a_{k_1}x^{k_1}+\cdots+a_{k_s}x^{k_s}$ with $n>k_1>\cdots>k_s$, $a_{k_i}\neq 0$, $\sqrt[n]{3}$ is a root of $f$, then by dividing both sides by $(\sqrt[n]{3})^{k_s}$, we may assume $b_1(\sqrt[n]{3})^{l_1}+\cdots+b_s=0$, with $l_1>l_2>\cdots>0$. Then how to derive a contradiction?
By Eisenstein's criterion applied to $f:=X^n-3$ with $p=3$ we see that this polynomial is irreducible over $\mathbb{Q}$ and has $\xi:=\sqrt[n]{3}$ as one of its roots. Thus $f$ is the minimal polynomial of $\xi$ over $\mathbb{Q}$.