There is weaker question Show that over any field $K$, such that $\mathbb Q \subset K$ the polynomial $x^3-3x+1$ is either irreducible or splits into linear factors. I tried to use the same method as follows:
Call $y_1,y_2,y_3$ the roots of $t^3−xt+1$ in $K$. You can compute $Δ=(y_1−y_2)^2(y_1−y_3)^2(y_2−y_3)^2$. It is symmetric when permuting the roots, so $Δ\in K$ (it is the discriminant of the polynomial).
Then, compute $δ′=(y_1−y_2)(y_1−y_3)\in K(y_1)$. If $Δ=δ_2$ with $δ\in K$, then $(y_2−y_3)^2=(\delta/\delta ' )^2$, and thus $(y_2−y_3)=±δ/δ′∈K(y_1)$. Combining this with $y_2+y_3=−y_1∈K(y_1)$, you obtain the expressions of $y_2$ and $y_3$ in $K(y_1)$.
The thing is, $Δ=4x^3-27$ is not always a square of some elements in $K$, so this arguement is not always true.
Another solution is as follows:
$f=t^3-xt+1=(t-a)(t^2+at+a^2-x)$. Assume $a\in K$, and another zero of $f$ ,$b$, contains in $K(a)$. Then $b=\alpha a^2+\beta a + \gamma$. Notes that $a^3=xa-1$, now solve equation $b^3 = xb-1$.
I think it may works, but the calculation is so complex.
Can you please give some suggestions?
Your general claim is false. For example, take $x=0$. Then $t^3+1$ has a root in $\Bbb{Q}$ itself but does not split there. That $-3$ is important. First it makes the polynomial irreducible over $\Bbb{Q}$. Second, it makes the discriminant be $4\cdot3^3-27 = 81$, which is a rational square, which implies that the Galois group over $\Bbb{Q}$ is $A_3$ so that the splitting field has degree $3$. That means adjoining a root, which gives a degree 3 extension within the splitting field, gives the entire splitting field and so splits the polynomial into linear factors. And since the polynomial is a cubic, if there is no root it remains irreducible, so in any extension it is irreducible or split into linear factors.