= Solution
The initial <displacement> lies in an <invariant subspace> of axial motions. Let the three opposite vertex pairs have positions
$$
\pm(1+q_x)\mathbf e_x,\qquad \pm(1+q_y)\mathbf e_y,\qquad \pm(1+q_z)\mathbf e_z.
$$
For a vertex on the positive $x$ axis, the two <springs> towards the $y$ vertices contribute $-k(q_x+q_y)$ in the $x$ direction; the two towards the $z$ vertices contribute $-k(q_x+q_z)$. Transverse components cancel in pairs. Cyclically,
$$
m\frac{d^2}{dt^2}\begin{pmatrix}q_x\\q_y\\q_z\end{pmatrix}
=-k\begin{pmatrix}2&1&1\\1&2&1\\1&1&2\end{pmatrix}
\begin{pmatrix}q_x\\q_y\\q_z\end{pmatrix}.
$$
The <matrix> is $I+\mathbf1\mathbf1^{\mathsf T}$. It has <eigenvalue> $4$ on the breathing <vector> $(1,1,1)$, and <eigenvalue> $1$ on the orthogonal plane $q_x+q_y+q_z=0$. This identifies the <axial normal modes of an octahedral spring framework> without a large characteristic <determinant>.
Put $\omega_0=\sqrt{k/m}$ and decompose the initial <vector> as
$$
(\epsilon_1,\epsilon_2,\epsilon_2)
=A(1,1,1)+D(2,-1,-1),\qquad
A=\frac{\epsilon_1+2\epsilon_2}{3},\quad D=\frac{\epsilon_1-\epsilon_2}{3}.
$$
Since all initial <velocities> are zero, only <cosine> time factors occur. The complete motion to first order in the <displacements> is
$$
\boxed{q_x(t)=A\cos(2\omega_0t)+2D\cos(\omega_0t),\qquad
q_y(t)=q_z(t)=A\cos(2\omega_0t)-D\cos(\omega_0t),}
$$
with the vertex positions given by the first displayed formula. No <translation> or <rotation> is excited: the initial <displacements> are orthogonal to those <zero modes> and their initial <velocities> vanish. The two checks $\epsilon_1=\epsilon_2$ and $\epsilon_1+2\epsilon_2=0$ respectively isolate pure breathing and pure traceless axial <oscillation>. Nonlinear corrections beyond the small-displacement approximation have been omitted.
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