I was trying to do some exercises to feel more confident about decomposition of reducible representations into direct sum of irreducible representations. The exercise involves decomposition of direct product of 3 $SU(3)$ vectors, and is the following:
$$3 \otimes 3 \otimes 3 = 3 \otimes (6 \oplus \bar{3}),$$
where $6$ represents the symmetric tensor and $\bar{3}$ represents the antisymmetric tensor.
Focusing on the $3 \otimes 6$ part:
$$ 3 \otimes 6 = q^iT^{\{jk\}}, $$
one should completely symmetrize and antisymmetrize the product into:
$$q^iT^{\{jk\}} = S^{\{ijk\}} + A^{[ijk]}.$$
This way we can show that these are the irreducible representations, and they transform within themselves under some $SU(N)$ transformations.
At this point my question is what is the explicit combination of $q^i$ and $T^{\{jk\}}$ that gives us $S^{\{ijk\}}$ and $A^{[ijk]}$?
Maybe it is redundant, but I want to emphasize what I try to obtain. So I show how I understand the first part of this calculation:
$$3 \otimes 3 = 6 \oplus \bar{3}$$ $$T^{ij} = \psi^i \phi^j$$ $$T^{ij} = S^{\{ij\}} + A^{[ij]}$$ $$S^{\{ij\}} = \frac{\psi^i \phi^j + \psi^j \phi^i}{2}$$ $$A^{[ij]} = \frac{\psi^i \phi^j - \psi^j \phi^i}{2} = \epsilon^{ijk}\epsilon_{klm}\psi^l \phi^m$$
