Let's have $SU(3)$ irreducible representations $3, \bar{3}$. How to get result that $$ 3\otimes 3 =6 \oplus \bar{3}? $$ I'm interested in $\bar{3}$ part. It's clear that for $3 \otimes 3$ we can use tensor rules by expanding corresponding matrix on symmetric $6$ and antisymmetric parts. But why we have $\bar{3}$, not $3$, for antisymmetric part?

Let's have $SU(3)$ irreducible representations $3, \bar{3}$. How to get result that $$ 3\otimes 3 =6 \oplus \bar{3}~? $$ I'm interested in $\bar{3}$ part. It's clear that for $3 \otimes 3$ we can use tensor rules by expanding corresponding matrix on symmetric $6$ and antisymmetric parts. But why we have $\bar{3}$, not $3$, for antisymmetric part?