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Let $\mathbf{r}$ be position relative to the center of mass, and let the new origin have vector $\mathbf{c}$ relative to it. The inertia tensor about this origin isExpanding and using $\int \mathbf{r}dm=0$ proves the tensor form of the parallel axis theorem:For a uniform cube of side $L$ and mass $M$, centered coordinate integrals give $I_0=(M{L}^2/6)I$. The displacement to a vertex is $(L/2,L/2,L/2)$ up to signs. Using axes along its incident edges givesThe principal moments of inertia are $\boxed{M{L}^2/6,11M{L}^2/12,11M{L}^2/12}$. The first principal axis is the body diagonal through the vertex; every direction perpendicular to it has the repeated moment.

First consider rotation predominantly about the third principal axis, writing ${\omega }_3=\Omega +O({\omega }_1^2+{\omega }_2^2)$. To first order the third Euler equation gives $\dot{\Omega }=0$, while the transverse components satisfyThereforeand the same equation holds for ${\omega }_2$.

Order the principal moments as $I_1<I_2<I_3$. The coefficient above is negative, so perturbations about axis three oscillate and remain bounded. Cyclically applying the same linear stability of principal-axis rotation calculation about axis one givesso that rotation is also stable. About axis two, however,which has an exponentially growing solution. Thus the intermediate axis theorem gives