= Explicit cubic SU3 tensor projections
{title2=$\mathbf3^{\otimes3}=\mathbf{10}\oplus\mathbf8\oplus\mathbf8\oplus\mathbf1$}
For $T^{abc}$, split its first pair into symmetric and antisymmetric parts. The fully symmetric part $S^{abc}$ is a <symmetric cubic representation of SU(3)>. Write $M^{abc}=T^{(ab)c}-S^{abc}$ and $O^a{}_d=\epsilon_{dbc}M^{abc}$. Then $O$ is traceless and $M^{abc}=(\epsilon^{bcd}O^a{}_d+\epsilon^{acd}O^b{}_d)/3$. The antisymmetric part is $T^{[ab]c}=\epsilon^{abd}N^c{}_d$, with $N^c{}_d=\epsilon_{dab}T^{abc}/2$. Splitting $N$ into its traceless part and trace gives the second <adjoint representation of SU(3)> and the singlet. These maps use <invariant tensors>, so they commute with the group action and explicitly realize the four irreducible summands.
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