= Solution
A field extension $L/K$ is a <splitting field> for $f\in K[X]$ when $f$ splits into linear factors over $L$ and $L$ is generated over $K$ by the roots of $f$. Equivalently, it is the smallest extension inside an algebraic closure over which $f$ splits.
To prove existence, choose an irreducible factor of $f$, adjoin one of its roots, and factor $f$ over the resulting finite extension. If a nonlinear factor remains, adjoin one of its roots and repeat. Each step lowers the total degree of the unsplit factors, so after finitely many finite extensions the polynomial splits. The field generated by the roots obtained is therefore a splitting field.
The precise uniqueness theorem says that if $L$ and $L'$ are splitting fields of the same polynomial $f$ over $K$, then there is a <field embedding> isomorphism
$$
\varphi:L\longrightarrow L'
$$
whose restriction to $K$ is the identity. More generally, an isomorphism of base fields carrying one polynomial to another extends to an isomorphism of their splitting fields. Thus splitting fields are unique up to base-field isomorphism; inside a fixed algebraic closure, the field generated by all the roots is literally unique. These facts are summarized by <existence and uniqueness of splitting fields>.
Let
$$
\alpha=\sqrt[3]{2}>0,
\qquad
\omega=e^{2\pi i/3}.
$$
The roots of $X^3-2$ are $\alpha,\omega\alpha,\omega^2\alpha$, so
$$
\boxed{L=\mathbb Q(\alpha,\omega)}
$$
is the <splitting field of x cubed minus two>. As a subfield of $\mathbb C$ it is unique, because any splitting field in $\mathbb C$ must contain all three roots and hence must contain both $\alpha$ and
$$
\omega=\frac{\omega\alpha}{\alpha};
$$
minimality then forces it to equal $\mathbb Q(\alpha,\omega)$.
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