Consider the family of pure cubic number fields $K = {\mathbb Q}(\sqrt[3]{m})$ for $m = a^3 \pm 3$.

**Proposition**. If $4 \mid a$ and $m$ is cubefree, then the
class number of $K$ is even.

**Proof.** Let $\omega = \sqrt[3]{m}$; the element $\alpha = a - \omega$
has norm $\pm 3$. Since $3$ is completely ramified, the element
$\varepsilon = \alpha^3/3$ is a unit.

If $m = a^3 + 3$, then $\varepsilon = 1 - 3a^2\omega + 3a\omega^2$. If $4 \mid m$, then $\varepsilon \equiv 1 \bmod 4$, hence $K(\sqrt{\varepsilon}\,)/K$ is an unramified quadratic extension.

Experiments seem to suggest that if $m = a^3+3$ and $a \equiv 2 \bmod 4$, then $h$ is also even, but there is no explanation, class field theoretic or otherwise. In fact, the class number is even for all cubefree values of $m$ for $a = 2, 4, \ldots, 2 \cdot 88$, but is odd for $a = 2 \cdot 89$.

This cannot be an accident; the parity of the class number in the case $m = a^3 - 3$ for $a \equiv 2 \bmod 4$ shows a more typical (i.e. more random) behaviour in that the class number is odd quite often.

Question: *How can this behaviour in the case $m = 8a^3+3$ be explained?*

My first guess would be that, for fields in this family, there is a family of ideals ${\mathfrak a}$ such that ${\mathfrak a}^2$ is principal, but I can't seem to find anything in this direction.

** Edit.** Dror's comment made me look at the family of elliptic curves $y^2 = x^3 - m$.
These have rank $\ge 1$, and by the parity conjecture rank $\ge 2$. An inequality due
to Billing now shows that $K$ has even class number. For details, see this
pdf file.
Actually, Paul Monsky stumbled across something similar for pure quartic fields;
see here.