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Can somebody explain in a simple way why, talking about representations $$3\otimes3\otimes3=1\oplus8\oplus8\oplus10~?$$

Here $3$ and $\bar{3}$ are the fundamental and anti-fundamental of $SU(3)$, in this case.

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Related $SU(3)$ post: physics.stackexchange.com/q/10403/2451 especially the answer physics.stackexchange.com/a/14586/2451 . Do you know Clebsch-Gordan decomposition of $SU(2)$ irreps? See e.g. physics.stackexchange.com/q/16098/2451 . –  Qmechanic Mar 21 '13 at 17:29
I found the group theory appendix (B) in Zee, Quantum Field Theory in a Nutshell, to be helpful for this stuff. –  Michael Brown Mar 22 '13 at 0:19
Removed subquestions that are duplicates in v3: physics.stackexchange.com/q/147243/2451 , physics.stackexchange.com/q/89173/2451 and links therein. –  Qmechanic Jun 19 at 21:02

1 Answer 1

\begin{equation} \boldsymbol{3}\boldsymbol{\otimes}\boldsymbol{3}\boldsymbol{\otimes}\boldsymbol{3}= \boldsymbol{1}\boldsymbol{\oplus}\boldsymbol{10}\boldsymbol{\oplus} \boldsymbol{8}^{\boldsymbol{\prime}}\boldsymbol{\oplus}\boldsymbol{8} \end{equation}

We talk about this because it explains the structure of a number of baryons in Particle Physics made from three quarks : 1 singlet - 1 decuplet - 2 octets, that is 27 baryons in total.
I refer to my answer in the following link for more details :


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