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Applying the Clebsch-Gordan decomposition for SU2 givesandOn the summand $V_{6-2j}$, interchange of the two tensor factors acts by $(-1{)}^j$; this is the flip parity in the SU2 tensor square. The even-$j$ summands make the symmetric square, so

Let $v_k=x^{n-k}{y}^k$. In the summand $V_{2n-2j}$ from part (d), a highest-weight vector isThe raising operator $x{\partial }_y$ annihilates this sum because consecutive terms cancel. Interchanging the tensor factors sends $w_j$ to $(-1{)}^j{w}_j$. Since each summand in part (d) has multiplicity one, the flip acts by that scalar on the whole summand. The alternating square therefore consists exactly of the odd-$j$ summands. This flip parity in the SU2 tensor square gives