With a little calculus is not hard to show this polynomial has three different real roots.
If $\,\alpha\,$ is one of the roots, then
$x^3-3x+1=(x-\alpha)\left(x^2+\alpha \,x+(\alpha^2-3)\right)$
The quadratic's discriminant is $\,\Delta:=12-3\alpha^2=3(2-\alpha)(2+\alpha)\,$
Now, again using a little a calculus, we can see the roots are in the following intervals:
$(-2,-1)\to\alpha_1\;\;\;\;\;\;\;(-1,1)\to\alpha_2\;\;\;\;\;\;(1,2)\to\alpha_3$
and since all the roots are in $\,(-2,2)\,$ we get that $\,\Delta>0\,$ no matter what root $\,\alpha\,$ we chose from the
three existing ones.
Thus, since the roots of the quadratic are
$\frac{-\alpha\pm\sqrt\Delta}{2}\in\Bbb Q(\alpha)$
We can see that $\,\Bbb Q(\alpha)\,$ contains them all and, thus, this is also true for any field containing $\,\alpha\,$.
Of course, if the polynomial has no roots in $\,\Bbb K\supset\Bbb Q\,$ then it must remain irreducible there (why?)