I read that any rank-2 tensor can be decomposed into the sum of a traceless symmetric tensor, an anti-symmetric tensor and a unit tensor, all closed under $SO(3)$. The three form an irreducible representation of $SO(3)$. The same is said to be possible for any types of tensors.
My question is: how a rank-3 tensor (and beyond) can be decomposed into parts that are closed under $SO(3)$?
You know the decomposition $3 \otimes 3 = 5 \oplus 3 \oplus 1$. The more general form is $n \otimes n = (2n - 1) \oplus (2n - 3) \oplus \ldots \oplus 3 \oplus 1$. So this pretty picture about how the sum of consecutive odd numbers is a square that you can find elsewhere on MSE has a surprising interpretation in terms of $SO(3)$-representations.
Now there is an even more general form. If I remember correctly it is:
$m \otimes n = (m + n - 1) \oplus (m + n - 3) \oplus \ldots \oplus (m -n + 3) \oplus (m -n + 1)$, assuming that $m \geq n$.
Now we can use this to compute $3 \otimes 3 \otimes 3$:
$$3 \otimes 3 \otimes 3 = 3 \otimes (5 \oplus 3 \oplus 1) = 5 \otimes 3 \oplus 3 \otimes 3 \oplus 3 \otimes 1 = (7 \oplus 5 \oplus 3) \oplus (5 \oplus 3 \oplus 1) \oplus 3 \\ = 7 \oplus 5 \oplus 5 \oplus 3 \oplus 3 \oplus 3 \oplus 1$$
Quick check: do these numbers add up to $27$? Yes. Ok, good.