= Solution
Let $X$ exchange the two equal-<spin> factors, so $X^\dagger=X=X^{-1}$. For the unperturbed <Hamiltonian operator>, <addition of angular momentum> gives
$$
E_J^{(0)}=\frac\alpha2\bigl[J(J+1)-2j(j+1)\bigr],\qquad J=0,1,\ldots,2j.
$$
For nonzero $\alpha$, different $J$ have distinct <energy eigenvalues>, while each <eigenspace> has the $2J+1$ values of $M$. The stated symmetry of the <Clebsch-Gordan coefficients> gives the <exchange parity of two coupled identical spins>:
$$
X|J,M\rangle=(-1)^{2j-J}|J,M\rangle.
$$
However, $D=(S_3^{(1)}-S_3^{(2)})/\hbar$ changes sign: $XDX=-D$. For any two <basis> vectors in the same $J$ <eigenspace>, insertion of $X^2=I$ therefore gives
$$
\langle J,M|D|J,M'\rangle=(-1)^{2(2j-J)}\langle J,M|XDX|J,M'\rangle=-\langle J,M|D|J,M'\rangle=0.
$$
Thus the entire projected perturbation $P_J\beta DP_J$ is zero, including its off-diagonal <matrix elements>. <Degenerate perturbation theory> now proves that every first-order-in-$\beta$ <energy> correction vanishes. This is <first-order cancellation for an exchange-odd perturbation>; checking only the individual diagonal expectations would not by itself settle a degenerate problem.
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