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Since a rank-3 tensor has 10 components and a rank-1 tensor has 3 components in $SU(3)$, I know that we are searching for the different irreducible representations of the tensor $v_{ijk}w_{l}$.

The fully symmetric part is equivalent to a rank-4 totally symmetric tensor with $v_{(ijk}w_{l)} = x_{ijkl}$. This will have 15 components.

But, I am not sure how to account for the rest 15. A possible setting includes a rank-2 tensor accounting for 6, rank-(1,1) accounting for 8 and one scalar using traceless property. But this is just me fixing the numbers without derivation.

$$\mathbf{10 \otimes 3 = 15 \ \oplus \ ?}$$

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    $\begingroup$ You can use Young tableaux for such decompositions. $\endgroup$ – G. Smith Feb 7 '20 at 18:17
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In tensor notation: $$v_{(ijk)}w_l = x_{(ijkl)} + \varepsilon_{ml(k} y_{ij)}^m + \underbrace{\varepsilon_{l(jk}z_{i)}}_{0} $$ $$ y^i_{(ij)} = 0 $$ Or: $$\mathbf{10 \otimes 3 = 15 \ \oplus \ 15^{\prime}}$$

Or by using Young tables:

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