Explicit cubic SU3 tensor projections ($3^{\otimes 3}=10\oplus 8\oplus 8\oplus 1$)

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 $O^a{}_d={ϵ}_{dbc}{M}^{abc}$. Then $O$ is traceless and $M^{abc}=({ϵ}^{bcd}{O}^a{}_d+{ϵ}^{acd}{O}^b{}_d)/3$. The antisymmetric part is $T^{[ab]c}={ϵ}^{abd}{N}^c{}_d$, with $N^c{}_d={ϵ}_{dab}{T}^{abc}/2$. 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.