= Solution
The <angular momentum commutation relations> are $[L_x,L_y]=i\hbar L_z$ and cyclic permutations. The product rule yields $[L_z,L_x^2]=i\hbar(L_yL_x+L_xL_y)$ and $[L_z,L_y^2]=-i\hbar(L_xL_y+L_yL_x)$, so \b[$[L_z,L^2]=0$]. Simultaneous <eigenstates> have <eigenvalues> $\hbar^2l(l+1)$ and $\hbar m$; <orbital angular momentum> has integer $l\ge0$ and $m=-l,\ldots,l$.
By cyclic <symmetry>, $L^2$ commutes with every component and hence every squared component. Therefore $[H,L^2]=0$: an <energy> eigenbasis can be chosen with definite $l$. For unequal transverse coefficients, $L_z$ need not commute with $H$, so definite $m$ is not generally available.
Expanding the <ladder operators> gives $L_+^2+L_-^2=2(L_x^2-L_y^2)$ and $L^2-L_z^2=L_x^2+L_y^2$. Comparing coefficients gives
$$
\boxed{\alpha=\frac{a_x-a_y}{4},\qquad\beta=\frac{a_x+a_y}{2},\qquad\gamma=a_z.}
$$
For $l=1,m=0$, both squared <ladder operators> vanish, since two raising or lowering steps would leave the multiplet. Thus $E_0=2\beta\hbar^2=\hbar^2(a_x+a_y)$. In the remaining basis $\{|1,1\rangle,|1,-1\rangle\}$ the <Hamiltonian> is
$$
\hbar^2\begin{pmatrix}\beta+\gamma&2\alpha\\2\alpha&\beta+\gamma\end{pmatrix}.
$$
Its symmetric and antisymmetric combinations have energies $\hbar^2(a_x+a_z)$ and $\hbar^2(a_y+a_z)$, respectively. Consequently
$$
\boxed{\{E_{l=1}\}=\hbar^2\{a_x+a_y,\ a_y+a_z,\ a_z+a_x\}.}
$$
The energies are related by permutation of the principal axes, not generally equal. Two coincide when the corresponding coefficients coincide; all three become $2a\hbar^2$ in the isotropic limit.
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