= Solution
Let $\rho_i=\rho_n=\rho$ and $\mu=\mu_{in}=\mu_{ni}$. Use <normal modes> proportional to $e^{i\mathbf k\cdot\mathbf x-i\omega t}$, so $\operatorname{Im}\omega<0$ means damping. The linear <displacements> satisfy
$$
\mathbf u_i=\partial_t\boldsymbol\xi_i=-i\omega\boldsymbol\xi_i,\qquad\mathbf u_n=\partial_t\boldsymbol\xi_n=-i\omega\boldsymbol\xi_n,\qquad\mathbf k\cdot\boldsymbol\xi_i=\mathbf k\cdot\boldsymbol\xi_n=0.
$$
For a nonzero-frequency perturbation, linearizing the <ideal magnetohydrodynamic induction equation> gives
$$
\partial_t\delta\mathbf B=B_0\partial_z\partial_t\boldsymbol\xi_i,\qquad\delta\mathbf B=B_0\partial_z\boldsymbol\xi_i=ik_\parallel B_0\boldsymbol\xi_i.
$$
Project both momentum equations with $\mathsf P=I-\mathbf k\mathbf k^T/k^2$. This removes the <pressure gradients>. The <magnetic tension> is already transverse to $\mathbf k$, and, with $a=k_\parallel v_A$, the projected equations are
$$
\ddot{\boldsymbol\xi}_i=-a^2\boldsymbol\xi_i-\mu(\dot{\boldsymbol\xi}_i-\dot{\boldsymbol\xi}_n),\qquad\ddot{\boldsymbol\xi}_n=-\mu(\dot{\boldsymbol\xi}_n-\dot{\boldsymbol\xi}_i).
$$
Here $v_A=B_0/\sqrt{4\pi\rho}$ uses the <ion> <density>, as stipulated. For either of the two independent transverse polarizations, the <equal-density ion-neutral Alfvén dispersion relation> follows from
$$
\begin{pmatrix}\omega^2-a^2+i\mu\omega&-i\mu\omega\\-i\mu\omega&\omega^2+i\mu\omega\end{pmatrix}\begin{pmatrix}\xi_i\\\xi_n\end{pmatrix}=0.
$$
Its determinant is $\omega[\omega^3+2i\mu\omega^2-a^2\omega-i\mu a^2]$. Removing the static neutral-displacement factor gives
$$
\boxed{\omega^3+2i\mu\omega^2-a^2\omega-i\mu a^2=0.}
$$
The removed factor is a time-independent neutral <displacement> with no <velocity> or magnetic perturbation, a relabeling of the homogeneous neutral fluid rather than a fourth physical <wave>. Direct elimination in the first-order system for $(u_i,u_n,\delta B)$ gives exactly the same cubic. This also handles the zero-frequency limits without introducing a displacement-label mode.
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