$\mathbb{Q}(x)$ is Galois over $\mathbb{Q}(x^{2})$ but not over $\mathbb{Q}(x^{3})$

Put \(t=x^2\). Then \(x\) satisfies \[ T^2-t\in\QQ(t)[T]=\QQ(x^2)[T]. \] Thus the degree is at most \(2\). It is not \(1\), because \(x\notin\QQ(x^2)\): every element of \(\QQ(x^2)\) is fixed by the automorphism of \(\QQ(x)\) sending \(x\) to \(-x\), whereas \(x\) is not. Hence \[ [\QQ(x):\QQ(x^2)]=2. \]

The minimal polynomial of \(x\) over \(\QQ(x^2)\) is \[ T^2-x^2, \] whose two roots in \(\QQ(x)\) are \(x\) and \(-x\). Since the characteristic is \(0\), the polynomial is separable, and it splits in \(\QQ(x)\). Hence the extension is Galois. Its two automorphisms are \[ x\longmapsto x, \qquad x\longmapsto -x, \] so the Galois group is cyclic of order \(2\).

Put \(s=x^3\). Then \(x\) satisfies \[ T^3-s\in\QQ(s)[T], \] so the degree is at most \(3\). The polynomial \(T^3-s\) is irreducible over \(\QQ(s)\) by Eisenstein’s criterion in \(\QQ[s][T]\) with the prime element \(s\): every nonleading coefficient is divisible by \(s\), the constant term \(-s\) is not divisible by \(s^2\), and the leading coefficient is not divisible by \(s\). By Gauss’s lemma it remains irreducible over \(\QQ(s)\). Therefore \[ [\QQ(x):\QQ(x^3)]=3. \]

The element \(\zeta_3\) is algebraic over \(\QQ\). We claim that every element of \(\QQ(x)\) algebraic over \(\QQ\) already lies in \(\QQ\).

Indeed, let \(r(x)\in\QQ(x)\) be nonconstant. Then \(x\) is algebraic over \(\QQ(r(x))\): writing \(r(x)=a(x)/b(x)\) with coprime \(a,b\in\QQ[x]\), the element \(x\) satisfies \[ a(T)-r(x)b(T)=0 \] over \(\QQ(r(x))\). Hence \(\QQ(x)/\QQ(r(x))\) is algebraic. Since \[ \operatorname{trdeg}_{\QQ}\QQ(x)=1, \] algebraic extensions preserve transcendence degree, so \[ \operatorname{trdeg}_{\QQ}\QQ(r(x))=1. \] Thus \(r(x)\) is transcendental over \(\QQ\). Therefore the only elements of \(\QQ(x)\) algebraic over \(\QQ\) are the constants in \(\QQ\).

Since \(\zeta_3\notin\QQ\), it follows that \(\zeta_3\notin\QQ(x)\).

Combine 2 and 5.