group theory - How to get result $3 otimes 3 = 6 oplus bar{3}$ for $SU(3)$ irreducible representations?

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?