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Construct the tensor-product weight diagram by placing one octet diagram at each triplet representation weight $u,d,s$ and adding multiplicities where the shifted diagrams overlap. Its highest weight isso remove the $(2,1)$ diagram by highest-weight character subtraction. Its unequal-sided hexagon has nine boundary representation weights of multiplicity one and three interior representation weights of multiplicity two, giving dimension $15$.

To see the remainder explicitly, measure diagram coordinates by $(a,b)=(2h_1,2\sqrt{3}{h}_2)$. After subtracting this $15$, the residual representation weights areThe next highest weight is $(h_1,h_2)=(0,2/\sqrt{3})=2{\omega }_2$, with Dynkin labels $(0,2)$. The conjugate sextet $\overline{6}$ is a triangular weight diagram containing each of these six representation weights once. Subtracting it leaves $(1,1),(-1,1),(0,-2)$ once each in the integer coordinates: precisely the triplet. The SU(3) triplet-octet decomposition is thereforeThe dimensions add to $15+6+3=24$, and the figure below verifies the equality at every representation weight, including overlapping interior representation weights.