How to calculate $3\otimes 3$ and $3\otimes 3\otimes 3$ in $SU(3)$? [closed]

EDIT: I have boiled my question down to

How many independent components does a rank three totally symmetric tensor have in $n$ dimensions?

A derivation would be nice too.

OP: I know that I can represent $3\otimes 3$ by a rank two tensor. Then the symmetric part is $3(3+1)/2=6$ and the antisymmetric part is $3(3-1)/2=3$. Thus $$3\otimes 3=6\oplus\bar 3$$ I'm not quite sure about the bar though...It is not clear to me why there is a bar when I do this diagrammatically, because the $\bar 3$ has the same isospin and hypercharge states as a quark triplet. So that's my first question.

So then $$3\otimes 3\otimes 3=3\otimes(6\oplus \bar{3})=(3\otimes 6)\oplus(3\otimes \bar 3)$$ I can represent $3\otimes \bar 3$ by a mixed tensor and separate out the trace. Thus $$3\otimes \bar 3=8\oplus 1$$ I get this just fine.

But I'm not sure how to go about doing $3\otimes 6$. Since $6$ is a symmetric tensor, $3\otimes 6$ should be a rank three tensor, symmetric over two indices. Then I could form a totally symmetric tensor and a tensor antisymmetric over two indices. But I have no clue which one is $10$ and which one is $8$ and obviously no clue how to prove it.

Any help would be greatly appreciated.

EDIT I: I once heard that in three dimensions an antisymmetric tensor is a vector. Is this the reason for writing the second rank antisymmetric tensor as $\bar{3}$? But how does one identify an antisymmetric tensor with the conjugate representation?

EDIT II: Strike the first question. If $\varphi^{kl}$ is the antisymmetric tensor, then I can use the Levi-Civita symbol to write $\varphi_i=\varepsilon_{ikl}\varphi^{kl}$, which is the $\bar{3}$.

EDIT III: I know how to get the $8$. Represent $6$ by the symmetric $S^{ij}$ and $3$ by the vector $\varphi^k$. Then define $\varphi_{lm}=\varepsilon_{lmk}\varphi^k$ and $T^{ij}_{lm}=S^{ij}\varphi_{lm}$. The tensor $T^{ij}_{lj}$ is traceless and is thus a $3^2-1=8$. From this I see that this $8$ has mixed symmetry properties, as is should, because $$T^{ij}_{lj}=T^{ji}_{lj}=-T^{ji}_{jl}=-T^{ij}_{jl}$$ Now I know that $6\otimes 3=10\oplus 8$, but how do I figure out that the $10$ is an irrep and totally symmetric?

EDIT IV: Now I tried writing $T^{ijk}=S^{(ik}\varphi^{k)}$. This should be the $10$. I wrote down the 27 components and verified there are $10$ degrees of freedom, as expected. But this is kinda laborious. So my question now becomes: How many independent components does a rank three totally symmetric tensor have in $n$ dimensions? If the answer is 10 for $n=3$, then my question will be answered.

closed as off-topic by Danu, Jim, bobie, Qmechanic♦Jan 6 '15 at 13:21

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