= Solution
Let $V=\mathbb C^3$ be the defining flavor <SU(3)> <representation>. Under $U\in SU(3)$, $q^\alpha\mapsto U^\alpha{}_{\beta}q^\beta$. The <antiquark> transforms in the <dual representation>:
$$
\bar q_\alpha\mapsto (U^{-1})^\beta{}_{\alpha}\bar q_\beta.
$$
Since $U$ is unitary, its lower-index column of components transforms by $U^*$, the complex conjugate numerical matrix. Thus $q^\alpha\bar q_\alpha$ is invariant. The defining triplet and its conjugate are inequivalent complex <representations>; conjugation reverses their <weights>. Write $\mathbf3=V$ and $\overline{\mathbf3}=V^*$, and distinguish a <direct sum> $\oplus$ from a <tensor product> $\otimes$.
First split $V\otimes V$ into <symmetric tensors> and <totally antisymmetric tensors>:
$$
V\otimes V=\operatorname{Sym}^2V\oplus\Lambda^2V.
$$
The symmetric part has dimension $3(3+1)/2=6$. Contracting the antisymmetric part with the invariant <Levi-Civita symbol> gives $A_\gamma=\tfrac12\epsilon_{\gamma\alpha\beta}A^{\alpha\beta}$, which transforms as a dual vector because $\det U=1$. Its inverse is $A^{\alpha\beta}=\epsilon^{\alpha\beta\gamma}A_\gamma$. Hence
$$
\boxed{\mathbf3\otimes\mathbf3=\overline{\mathbf3}\oplus\mathbf6.}
$$
The symmetric part is the irreducible <highest-weight representation> with <Dynkin labels> $(2,0)$; its six monomials give the standard symmetric-square <representation>.
For three triplets, the antisymmetric first-pair sector becomes $V^*\otimes V$. Its <tensors> $M^\alpha{}_{\beta}$ split into their scalar trace and traceless part:
$$
V^*\otimes V=\mathbf1\oplus\mathbf8,\qquad
M=\frac{\operatorname{tr}M}{3}\mathbf1+\left(M-\frac{\operatorname{tr}M}{3}\mathbf1\right).
$$
The traceless part is the <adjoint representation of SU(3)>; conjugation acts as $M\mapsto UMU^{-1}$. It is irreducible because an invariant subspace would be an ideal in the simple complex <special linear Lie algebra> $\mathfrak{sl}_3$.
The symmetric first-pair sector consists of $T^{abc}=T^{bac}$ and has dimension $6\cdot3=18$. Complete symmetrization maps it onto $\operatorname{Sym}^3V$, whose dimension is $\binom53=10$. Its kernel is the mixed-symmetry eight-dimensional sector. More explicitly, the equivariant map $T^{abc}\mapsto M^a{}_e=\epsilon_{ebc}T^{abc}$ has trace zero because the first two indices of $T$ are symmetric; it annihilates the completely symmetric part and is not identically zero. Since the target adjoint module is irreducible, the map is onto its eight-dimensional space. Its kernel is consequently exactly the ten-dimensional completely symmetric part. Complete symmetrization supplies an invariant splitting, so this sector is $\mathbf{10}\oplus\mathbf8$. Combining the sectors gives the <tensor cube of the defining SU(3) representation>
$$
\boxed{\mathbf3^{\otimes3}=\mathbf1\oplus\mathbf8\oplus\mathbf8\oplus\mathbf{10}.}
$$
Here the decuplet has <Dynkin labels> $(3,0)$, and the singlet is the fully antisymmetric contraction $\epsilon_{abc}T^{abc}$. The two octets are distinct copies, not two different irreducible types.
For <two conjugate SU(3) triplets with a triplet>, begin with $T^a{}_{bc}$ and split its lower pair. An antisymmetric lower pair is identified by $\epsilon^{bcd}$ with an upper index, giving $V\otimes V=\mathbf6\oplus\overline{\mathbf3}$. The symmetric lower-pair sector has dimension $3\cdot6=18$, and its contraction is the equivariant map
$$
C(T)_b=\sum_aT^a{}_{ba}.
$$
It is onto $V^*$, since the invariant embedding
$$
\iota(v)^a{}_{bc}=\frac14(\delta^a_bv_c+\delta^a_cv_b)
$$
satisfies $C\iota(v)=v$. Every such <tensor> therefore splits uniquely as $\iota(C(T))+(T-\iota(C(T)))$. The latter <tensor> is symmetric in its lower indices and has zero contraction, with dimension $18-3=15$. It contains the <highest-weight vector> $e_1\otimes(e_3^*)^2$ of <Dynkin labels> $(1,2)$; contraction vanishes because its upper and lower indices differ. The <Weyl dimension formula> for an <SU(3)> module with labels $(p,q)$ is $\dim V_{(p,q)}=(p+1)(q+1)(p+q+2)/2$, which gives $15$ for $(1,2)$. These <tensor> <representations> are unitary, so invariant subspaces have invariant orthogonal complements; the irreducible highest-<weight> summand containing this vector exhausts the fifteen-dimensional kernel. Thus
$$
\boxed{\overline{\mathbf3}\otimes\overline{\mathbf3}\otimes\mathbf3
=\overline{\mathbf3}\oplus\overline{\mathbf3}\oplus\mathbf6\oplus V_{(1,2)}.}
$$
The question's dimension label $15$ denotes $V_{(1,2)}$, rather than its conjugate $V_{(2,1)}$; they have equal dimensions but different <tensors>. Conjugating the triplet-cube decomposition gives
$$
\boxed{\overline{\mathbf3}^{\otimes3}=\mathbf1\oplus\mathbf8\oplus\mathbf8\oplus\overline{\mathbf{10}}.}
$$
The octet is self-conjugate, whereas the symmetric decuplet changes $(3,0)$ to $(0,3)$.
Now consider the flavor <diquark> $D_\alpha\propto\epsilon_{\alpha\beta\gamma}q^\beta q^\gamma$. This notation describes an antisymmetric two-<quark> flavor state; color and <spin> factors are separate. In particular $D_3=(ud-du)/\sqrt2$ has <isospin> zero, <baryon number> $2/3$, <strangeness> zero and <flavor hypercharge> $2/3$. The strange <antiquark> has $I=0$, $B=-1/3$, $S=+1$ and $Y=2/3$. Thus $D_3D_3\bar s$ has
$$
\boxed{B=1,\qquad S=+1,\qquad Y=2,\qquad I=0,\qquad Q=+1.}
$$
The charge follows from the <Gell-Mann--Nishijima formula>. Each of the three factors has lower flavor index $3$, so their product is the completely symmetric <tensor> component $A_{333}$. It lies entirely in $\operatorname{Sym}^3V^*=\overline{\mathbf{10}}$, the <pentaquark antidecuplet>. Equivalently, the symmetric diquark pair belongs to $\overline{\mathbf6}$, and $\overline{\mathbf6}\otimes\overline{\mathbf3}=\overline{\mathbf{10}}\oplus\mathbf8$, with this highest-hypercharge state in the antidecuplet.
It cannot be in an octet. The octet has the <weights> of a traceless $q^\alpha\bar q_\beta$ <tensor>, whose hypercharges are $Y_\alpha-Y_\beta$. For $(u,d,s)$ the <quark> values are $1/3,1/3,-2/3$, so these differences are only $0,+1,-1$; there is no $Y=2$ <weight>. The octet's top row instead has $Y=1$ and $I=1/2$. This is a flavor-<representation> assignment for a possible <pentaquark>, not a proof that the configuration binds or that its full color-<spin>-orbital wavefunction satisfies all <Fermi statistics> constraints.
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