= Two conjugate SU(3) triplets with a triplet
{title2=$\overline{\mathbf3}\otimes\overline{\mathbf3}\otimes\mathbf3=2\overline{\mathbf3}\oplus\mathbf6\oplus V_{(1,2)}$}
Split the two dual factors into antisymmetric and symmetric <tensors>. The invariant <Levi-Civita symbol> identifies the antisymmetric pair with $\mathbf3$, giving $\mathbf6\oplus\overline{\mathbf3}$ after multiplying by the remaining triplet. On the symmetric part, contraction $T^a{}_{ba}$ maps onto $\overline{\mathbf3}$; the equivariant splitting is $T^a{}_{bc}=(\delta^a_bv_c+\delta^a_cv_b)/4$. Its trace-free kernel has dimension $18-3=15$ and highest weight with <Dynkin labels> $(1,2)$. The <Weyl dimension formula> gives dimension $15$, so this is exactly the irreducible kernel. It is conjugate to the other fifteen-dimensional module with labels $(2,1)$; dimension alone does not distinguish the two.
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