standard model - Why isn't there a second baryon octet?

Let's temporarily ignore spin. If 3 denotes the standard representation of $SU(3)_F$, 1 the trivial rep, 8 the adjoint rep and 10 the symmetric cube, then it's well-known that

$$3 \otimes 3 \otimes 3 = 1 \oplus 8 \oplus 8 \oplus 10.$$

Interpreting the 3's as the space of up/down/strange flavour states for a quark, the tensor cube is interpreted as the space of baryon states that can be obtained by combining three light quarks. Obviously spin matters, but at least this should give a classification of baryons modulo spin into $SU(3)_F$-multiplets.

There are two octets here, but in the literature I have only seen one of them described. What is the second octet?

I appreciate that the answer may be "it's more complicated than that".