= Solution
A ground-state <baryon> contains three <quarks> in a spatially symmetric $L=0$ <wavefunction>, giving positive <parity>. Each <quark> is a <spin one-half> <fermion>, so the complete <wavefunction> must be antisymmetric under interchange of its complete quark labels. A physical <baryon> is a <colour singlet>, with the antisymmetric colour factor
$$
\psi_{\mathrm{colour}}\propto\epsilon_{abc}|a\,b\,c\rangle.
$$
Consequently the combined spin-and-flavour factor must be symmetric. The product notation in the question organizes the degrees of freedom; it need not mean that spin and flavour are separately unentangled. In particular the octet requires a correlated sum of products.
For the three light flavours, the <flavour symmetry> decomposition is
$$
\mathbf3\otimes\mathbf3\otimes\mathbf3
=\mathbf{10}_{S}\oplus\mathbf8_{M}\oplus\mathbf8_{M}\oplus\mathbf1_{A}.
$$
The subscripts describe permutation symmetry, with $M$ denoting mixed symmetry. The <addition of angular momentum> of three <spin one-half> particles gives
$$
\mathbf2\otimes\mathbf2\otimes\mathbf2
=\mathbf4_S\oplus\mathbf2_M\oplus\mathbf2_M.
$$
The symmetric spin quartet has total $J=3/2$; the two spin doublets have $J=1/2$.
For $J=3/2$, the spin factor is symmetric, so the flavour factor must also be symmetric: this gives the \b[spin-$3/2$ baryon decuplet]. Its flavour content is four $\Delta$ states, three $\Sigma^*$ states, two $\Xi^*$ states, and one $\Omega$ state.
For $J=1/2$, the mixed-symmetry spin functions combine with the mixed-symmetry flavour functions to make one fully symmetric spin-flavour factor for each octet weight: this gives the \b[spin-$1/2$ baryon octet], containing two <nucleons>, three <Sigma baryons>, one <Lambda baryon>, and two <Xi baryons>. The two mixed-symmetry octets in the raw flavour product are not two distinct ground-state octets; their permutation partners are correlated with the spin factors. An antisymmetric flavour singlet would require a fully antisymmetric spin factor to make a symmetric product, but three spin states drawn from a two-dimensional single-quark spin space have no such factor.
Equivalently, the symmetric three-particle representation of <spin-flavour symmetry> has dimension $\binom{8}{3}=56$ and decomposes under flavour and spin into
$$
\boxed{\mathbf{56}_{S}=(\mathbf{10},\mathbf4)\oplus(\mathbf8,\mathbf2),
\qquad56=10\times4+8\times2.}
$$
These are approximate multiplets: the larger <strange quark> <mass> breaks exact <flavour symmetry> and splits their <energies>.
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