Is any rational extension of degree 3 of the form $\mathbb{Q}(\sqrt[3]{d})$?

Question

If $K/\mathbb{Q}$ is an extension of degree $2$, then necessarily it is of the form $\mathbb{Q}(\sqrt{d})$. So, I am wondering if a similar philosophy holds for extension over $\mathbb{Q}$ of degree 3.

Answer 1

Your answer I think should be negative because if, for example, $\theta$ is the real root of $x^3+x+1=0$ then $K=\mathbb Q(\theta)$ is of degree $3$ but $\theta\approx -0.6823$ being such that $\theta^3+\theta+1=0$ is not of the form you asked.

Answer 2

$\mathbb{Q}(\sqrt[3]{d})$ is not a Galois extension if $d$ is not a power of $3$. In fact, it is not normal, there exists Galois extensions of degree $3$,therefore they cannot be isomorphic to $\mathbb({Q}\sqrt[3]{d})$.

Finding a Galois extension of $\Bbb Q$ of degree $3$