= Solution
Label the equilibrium vertices in the order $+x,-x,+y,-y,+z,-z$, and write their <displacements> as $\mathbf u_i(t)\in\mathbb R^3$. Their eighteen Cartesian components are <generalized coordinates>. Linearize the axial Hookean <springs> about the unstressed equilibrium. For an edge with unit direction $\mathbf n_{ij}$, its first-order extension is $\mathbf n_{ij}\cdot(\mathbf u_i-\mathbf u_j)$, so the quadratic <potential energy> is
$$
V=\frac k2\sum_{\langle ij\rangle}[\mathbf n_{ij}\cdot(\mathbf u_i-\mathbf u_j)]^2.
$$
Three <zero modes> are <translations>: for each Cartesian <unit vector> $\mathbf e_\alpha$, take $\mathbf u_i=\mathbf e_\alpha$ at every vertex. Three more are infinitesimal <rotations> $\mathbf u_i=\mathbf e_\alpha\times\mathbf r_i$. Explicitly, in the stated vertex order, their <displacement> <vectors> are
$$
R_x:\quad (\mathbf0,\mathbf0,\mathbf e_z,-\mathbf e_z,-\mathbf e_y,\mathbf e_y),
$$
$$
R_y:\quad (-\mathbf e_z,\mathbf e_z,\mathbf0,\mathbf0,\mathbf e_x,-\mathbf e_x),
$$
$$
R_z:\quad (\mathbf e_y,-\mathbf e_y,-\mathbf e_x,\mathbf e_x,\mathbf0,\mathbf0).
$$
<Translations> leave each <displacement> difference zero. For <rotations>,
$$
\mathbf n_{ij}\cdot[\boldsymbol\Omega\times(\mathbf r_i-\mathbf r_j)]=0
$$
because the equilibrium edge <vector> is parallel to $\mathbf n_{ij}$. Hence every <spring> has zero first-order extension in all six patterns, and the <stiffness matrix> annihilates them.
The <translations> have nonzero summed <displacement>, while all the <rotations> have zero sum. The three rotational patterns are themselves independent because the vertices do not lie on a single line. Thus
$$
\boxed{\text{there are at least six independent zero modes: three translations and three rotations}.}
$$
These are the <rigid-motion zero modes of a spring framework>. The rotational <displacements> describe infinitesimal rigid <rotations>; an arbitrary finite <rotation> would require its nonlinear <displacement> field.
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