$x^n - a$ is irreducible over $\mathbb{Q}$?
Use unique factorization in $\Bbb{C}[x]$. You know that the zeros of the polynomial in $\Bbb{C}$ are $x_k=\zeta^ka^{1/n},$ $\zeta=e^{2\pi i/n}$, $k=0,1,2,\ldots,n-1$. If $f(x)\in\Bbb{Q}[x]$ were a factor of $x^n-a$, then we have $$ f(x)=\prod_{k\in S}(x-x_k) $$ for some subset $S\subset\{0,1,2,\ldots,k-1\}$.
What can you say about the constant term of $f(x)$? Remember that it should be rational, and thus have a rational absolute value.