 Solution

The nuclear quadrupole vibration has five one-phonon magnetic substates, $m=-2,-1,0,1,2$. Bose-Einstein statistics restrict three identical phonons to the symmetric power $S^3(V_2)$, of dimension $\left(\genfrac{}{}{0ex}{}{5+3-1}{3}\right)=35$. Count occupation states $(n_{-2},\dots ,n_2)$ with $\sum n_m=3$ and $\sum m{n}_m=M$. For nonnegative $M$ the counts areEach spin-$J$ irreducible representation contributes one state at every projection from $-J$ to $J$. magnetic-substate subtraction therefore gives multiplicity $d_J-d_{J+1}$: one each at $J=0,2,3,4,6$, and none at $1,5$. Thus the three-quadrupole-phonon multiplet isThe dimension check is $1+5+7+9+13=35$. The parity is positive because each constituent phonon has positive parity. Pairwise angular-momentum triangle rules alone would include states incompatible with complete boson symmetrization.