Let $\alpha$ denote a root of $p = x^3 - x - 4$ and let $K = \mathbb{Q}(\alpha)$. Find the ring of integers of $K$, and show that $K$ has class number $1$.
If we compute the Minkowski bound, then we realize that we only need to analyze the prime lying over $2$, $\mathfrak{p}_2$, and based on how $p$ factors $\mod 2$ we can say that the class group is generated by $\mathfrak{p}_2$. From here, I do not know how to ensure that $K$ has class number $1$.
If I am not mistaken, the minimal polynomial of $\beta=\dfrac{\alpha+\alpha^2}{2}$ is $q=X^3-X^2-3X-2$.
Assuming that your computation of the ring of integers is correct (and it is!), we have $(2)=(2,\beta)(2,\beta^2-\beta-1)$ (note that the decomposition of primes are reflected by the decomposition of $q$ modulo primes, not by the decomposition of $p$ !!!!)
Minkowski bound is $<3$, so it indeed remains to prove that $\mathfrak{p}_{2}=(2,\beta)$ is principal.
But $2=3\beta+\beta^2+\beta^3=\beta(3+\beta+\beta^2)\in\mathfrak{p}_2$, so $\mathfrak{p}_2=(\beta)$ is principal.