= Solution
In a <tensor product of Lie algebra representations>, the Cartan generators act as sums on the tensor factors. Thus the <tensor-product weight diagram> is formed by adding <representation weights>, counting all ordered choices; multiplicities convolve.
For three triplets, the three vertices $3u,3d,3s$ each arise once. Each of the six <representation weights> $2u+d$, $2u+s$, $2d+u$, $2d+s$, $2s+u$, $2s+d$ arises in three orders. The central <representation weight> $u+d+s=0$ arises in six orders. The resulting <weight diagram> has total <dimension> $3+6\cdot3+6=27$.
Its maximal <highest weight> is $3u=3\omega_1$, with <Dynkin labels> $(3,0)$. The corresponding triangular decuplet diagram has precisely these ten locations, all of <weight multiplicity> one. Subtract it by <highest-weight character subtraction>. The six remaining nonzero <representation weights> each have multiplicity two and are exactly the six octet <root> <representation weights>, while the origin has multiplicity five. Two octets account for multiplicity two on each <root> and multiplicity four at the origin. The remaining origin is one singlet. Therefore the <tensor cube of the defining SU(3) representation> decomposes as
$$
\boxed{\mathbf3\otimes\mathbf3\otimes\mathbf3
=\mathbf{10}_{(3,0)}\oplus
\mathbf8_{(1,1)}\oplus\mathbf8_{(1,1)}\oplus\mathbf1_{(0,0)}.}
$$
The <dimension> check is $10+8+8+1=27$. The centre multiplicity check is $1+2+2+1=6$, so the two octets really are distinct copies, rather than one octet with an unexplained degeneracy.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/iii/paper-50-su3-tensor-cube.png]
{title=Weight-by-weight decomposition of three SU(3) triplets}
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