Tensor square of the SU(3) adjoint representation ($8\otimes 8=1\oplus 8_S\oplus 8_A\oplus 10\oplus \overline{10}\oplus 27$)

For traceless matrices $T,S$ transforming by conjugation under SU(3), the trace $\mathrm{tr}(TS)$ supplies a singlet, the commutator $[T,S]$ an antisymmetric octet, and the traceless anticommutator a symmetric octet. Symmetrizing the two upper and two lower indices and removing traces supplies the 27. Contracting an antisymmetric lower pair with the invariant alternating tensor and symmetrizing the remaining three upper indices supplies the 10, and the conjugate construction supplies the conjugate 10. These nonzero equivariant maps exhaust the dimensions: the symmetric part has $36=1+8+27$, the antisymmetric part $28=8+10+10$.