# Simple/elementary explanation for $\mathbf{3} \otimes \mathbf{\bar{3}} = \mathbf{8} \oplus \mathbf{1}$? [duplicate]

I am preparing a talk on the Eightfold Way, and am attempting to explain the spectra of the light mesons/baryons via representation theory. It will be delivered to students who have never seen representation theory before. I understand the arrangement of particles can be explained by reps of $$\mathrm{SU}(3)$$, e.g. the light mesons ($$\mathrm{u}$$, $$\mathrm{d}$$, $$\mathrm{s}$$) correspond to the representation $$\mathbf{3} \otimes \mathbf{\bar{3}}$$ of $$\mathrm{SU}(3)$$, which I understand decomposes as $$\mathbf{3} \otimes \mathbf{\bar{3}} = \mathbf{8} \oplus \mathbf{1}$$.

Is there a simple and elementary way to derive this decomposition, i.e. without needing to know a lot of Lie groups/rep theory?

I've seen Young diagrams used as a tool, but still have not been able to understand how they work. If someone could give a self-contained explanation, or direct me to where I could find one, that would be great. I know group theory and surface-level stuff, and am willing to blindly accept some facts [e.g. that there exist representations $$\mathbf{1}$$, $$\mathbf{3}$$, $$\mathbf{6}$$, $$\mathbf{8}$$ of $$\mathrm{SU}(3)$$], but I'd like to give some motivation for the formula $$\mathbf{3} \otimes \mathbf{\bar{3}} = \mathbf{8} \oplus \mathbf{1}$$ without just handwaving it.

Same for the baryons: $$\mathbf{3} \otimes \mathbf{3} \otimes \mathbf{3} = \mathbf{10} \oplus \mathbf{8} \oplus \mathbf{8} \oplus \mathbf{1}$$.