= Solution
Since $X=L(a+a^\dagger)$, the perturbation is
$$
\lambda\hbar\omega(a+a^\dagger)^3.
$$
Every diagonal matrix element vanishes by <parity>, so all first-order shifts are zero. For the ground state,
$$
(a+a^\dagger)^3|0\rangle=3|1\rangle+\sqrt6|3\rangle.
$$
The second-order shift is therefore
$$
\lambda^2(\hbar\omega)^2
\left(\frac9{-\hbar\omega}+\frac6{-3\hbar\omega}\right)
=-11\lambda^2\hbar\omega.
$$
Thus
$$
\boxed{E_0=\hbar\omega\left(\frac12-11\lambda^2\right)+O(\lambda^3)}.
$$
For every nonzero real $\lambda$, the cubic potential is unbounded below on one side, so there are no genuine bound-state eigenvalues continuously connected to the harmonic levels. The formal perturbation series does not converge; it is at best asymptotic to resonance energies.
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