= Solution
The uniform planar eight-electron ring has a partially filled frontier degeneracy. <Bond-length alternation> splits those levels, permitting the two frontier <electrons> to occupy the lowered orbital and opening a gap to the empty partner. This is the electronic mechanism for destabilizing the equal-bond planar geometry.
A controlled comparison keeps the mean resonance integral fixed. Put
$$
\beta_S=\beta(1+\delta),\qquad
\beta_L=\beta(1-\delta),\qquad \beta<0,\quad0<\delta<1.
$$
Using part (i), the change from the uniform ring is
$$
\Delta E_\pi=4\beta\delta
+4\sqrt2\,\beta\big(\sqrt{1+\delta^2}-1\big)
=4\beta\delta+2\sqrt2\,\beta\delta^2+O(\delta^4)<0.
$$
Thus \b[a small alternation lowers the electronic <energy>], with a leading gain linear in $\delta$. An ordinary elastic cost starts quadratically, so the equal-bond planar state is unstable in this model. The equilibrium distortion requires those geometric and sigma-bond costs as well.
The eight $\pi$ <electrons> constitute a $4n$ cyclic count rather than the closed-shell $4n+2$ case associated with <aromaticity>. The result concerns the imposed planar geometry; it does not establish the unconstrained three-dimensional equilibrium shape of <cyclooctatetraene>.
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