= Solution
The <Kronecker delta> is the identity map on the defining representation. Its transformation is $A^\alpha{}_\gamma\delta^\gamma{}_\delta(A^{-1})^\delta{}_\beta=\delta^\alpha{}_\beta$. For the alternating <Levi-Civita symbol>, the determinant identity gives
$$
A^\alpha{}_aA^\beta{}_bA^\gamma{}_c\epsilon^{abc}=(\det A)\epsilon^{\alpha\beta\gamma}=\epsilon^{\alpha\beta\gamma}.
$$
The lower-index alternating <tensor> transforms with three inverse <matrices> and acquires $\det(A^{-1})=1$ instead. Thus \b[$\delta^\alpha{}_\beta$, $\epsilon^{\alpha\beta\gamma}$ and $\epsilon_{\alpha\beta\gamma}$ are invariant <tensors>] of <SU(3)>. These identities use unimodularity; unitarity identifies the dual representation with the complex-conjugate defining representation.
For a <mixed SU(3) tensor representation> $T^a{}_b$, the <matrix trace> is invariant. Hence
$$
T^a{}_b=\frac13\delta^a{}_b\,T^c{}_c+Q^a{}_b,\qquad Q^a{}_a=0,
$$
gives
$$
\boxed{\mathbf3\otimes\overline{\mathbf3}=\mathbf1\oplus\mathbf8.}
$$
The trace is the singlet. The traceless <matrix> transforms by conjugation, so it is the <adjoint representation of SU(3)>. It is irreducible: an invariant complex subspace is invariant under commutators with the complexified algebra $\mathfrak{sl}_3$, and would be an ideal in that simple <Lie algebra>. Only the zero and full traceless-<matrix> spaces are possible.
For the cubic <tensor>, define its symmetric and antisymmetric first-pair parts by
$$
T_+^{abc}=\frac12(T^{abc}+T^{bac}),\qquad
T_-^{abc}=\frac12(T^{abc}-T^{bac}),
$$
and let $S^{abc}=T^{(abc)}$ be its completely symmetric part. This has $\binom{5}{3}=10$ independent components and transforms as the <symmetric cubic representation of SU(3)>. Its irreducibility can be seen in the homogeneous cubic-polynomial realization: the simultaneous highest-weight conditions $z_1\partial_{z_2}P=0$ and $z_2\partial_{z_3}P=0$ leave only $P\propto z_1^3$. A finite-dimensional unitary representation is completely reducible, and each irreducible summand has a <highest-weight vector>, so this single highest-weight line implies a single irreducible summand. Lowering generates all ten monomials.
Remove this fully symmetric part from $T_+$, putting $M^{abc}=T_+^{abc}-S^{abc}$. Define
$$
O^a{}_d=\epsilon_{dbc}M^{abc}.
$$
Its trace vanishes because $M$ is symmetric in its first two indices. Invariance of $\epsilon$ makes $O$ transform as one upper and one lower index. The inverse map is
$$
M^{abc}=\frac13\left(\epsilon^{bcd}O^a{}_d+\epsilon^{acd}O^b{}_d\right).
$$
Indeed contraction with $\epsilon_{ebc}$ gives $(2O^a{}_e+O^a{}_e-\delta^a{}_e\operatorname{tr}O)/3=O^a{}_e$. The inverse has the required first-pair symmetry and zero completely symmetric part. Thus this eight-dimensional remainder is another <adjoint representation of SU(3)>.
For the antisymmetric part, define
$$
N^c{}_d=\frac12\epsilon_{dab}T_-^{abc},\qquad
T_-^{abc}=\epsilon^{abd}N^c{}_d.
$$
Now decompose $N^c{}_d=U^c{}_d+\delta^c{}_d\operatorname{tr}N/3$, with $U$ traceless. The first term supplies a second independent octet; the trace supplies a singlet proportional to $\epsilon^{abc}$. With $q=\epsilon_{abc}T^{abc}/6=\operatorname{tr}N/3$, the <explicit cubic SU3 tensor projections> reconstruct the entire <tensor>:
$$
T^{abc}=S^{abc}
+\frac13\left(\epsilon^{bcd}O^a{}_d+\epsilon^{acd}O^b{}_d\right)
+\epsilon^{abd}U^c{}_d+q\epsilon^{abc}.
$$
All projections commute with the <SU(3)> action because they use permutations and <invariant tensors>. Their inverse maps and the dimension count $10+8+8+1=27$ establish
$$
\boxed{\mathbf3\otimes\mathbf3\otimes\mathbf3=\mathbf{10}\oplus\mathbf8\oplus\mathbf8\oplus\mathbf1.}
$$
The two octets are equivalent as group representations but distinct subspaces in this decomposition; one comes from the first-pair symmetric sector and the other from the first-pair antisymmetric sector. General permutations can mix these equivalent copies.
In the <quark model>, the <up quark>, <down quark> and <strange quark> form the defining triplet of approximate <flavor symmetry>. Three-quark <baryons> therefore have precisely these possible flavor representations. The fully symmetric decuplet corresponds to the observed spin-$3/2$ <baryon decuplet>: the four <Delta baryons>, three $\Sigma^*$ states, two $\Xi^*$ states and the <Omega baryon>. Their strangeness rows have <flavor hypercharge> and <isospin> $(Y,I)=(1,3/2),(0,1),(-1,1/2),(-2,0)$; $uuu$ and $sss$ occupy the extreme $\Delta^{++}$ and $\Omega^-$ weights.
The observed spin-$1/2$ <baryon octet> contains the <nucleons>, three <Sigma baryons>, the <Lambda baryon> and two <Xi baryons>. In its $Y=0$ row, $\Sigma^0$ and $\Lambda$ share the same third-component/hypercharge weight but have different total <isospin>. Approximate flavor symmetry explains multiplet organization, not exact equality of all their masses: the strange-quark mass and electromagnetic interactions break it.
The two octet copies in the flavor <tensor> do not require two identical ground-state octets. A <three-quark colour singlet> is antisymmetric in color; for a symmetric ground-state orbital wavefunction, the <Pauli exclusion principle> requires a symmetric combined spin-flavor state. The mixed flavor octet <tensors> combine with the mixed spin-$1/2$ <tensors> to supply that symmetry. The symmetric flavor decuplet combines with symmetric spin $3/2$. Together they form the <symmetric spin-flavour SU6 representation>, with $10\times4+8\times2=56$ spin-flavor states. A completely antisymmetric flavor singlet cannot appear in this spatial ground state because three spin-$1/2$ indices have no fully antisymmetric spin state: $\bigwedge^3\mathbb C^2=0$. <Three-quark flavour singlet> configurations instead require a suitable excited spatial/spin symmetry. Flavor <SU(3)> here is distinct from the exact color gauge symmetry; treating them as the same group action would misidentify the physical multiplets.
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