Decomposition of $SU(3)$ representation $6\times 15$ into irreducibles? The 6 and 15 dimensional representations of $SU(3)$ are irreducible. The 90 dimensional tensor product representation $6\times 15$ decomposes into a sum of irreducible representations. What factors occur and with what multiplicity?
Note: by 6 I mean the 2 index symmetric representation and not its complex conjugate (which is also 6 dimensional). Similarly for 15.
 A: This question is borderline between what is on topic and what isn't; if you want to do a number of computations like this you should pick up a book on representation theory. My standard recommendations for $SL_n$ rep theory are Chapter 8 of Fulton's Young Tableaux or Appendix II (by Fomin) in Stanley's Enumerative Combinatorics Volume 2. You then just need someone to tell you that the finite dimensional representation theories of $SU$ and $SL$ are the same. Perhaps someone will recommned a book that works directly in $SU$.
But I can easily imagine someone working in some other area just needing one answer and not wanting to open the book, so here is the computation.
Irreducible $SU$ representations are indexed by partitions. The $6$ and $15$ dimensional representations you speak of are $\mathrm{Sym}^2$ and $\mathrm{Sym}^4$ of the standard representation, so they are represented by the partitions $(2)$ and $(4)$. Tensor products where one factor is a single horizontal row are computed by the Pieri rule. There are three components, each of multiplicity $1$, corresponding to the partitions $(6)$, $(5,1)$ and $(4,2)$. The first one is $\mathrm{Sym}^6$, with dimension $28$. 
The other two don't have simple descriptions, but their characters are the Schur functions
$$s_{51}(x,y,z) = x^5 y + x^4 y^2 + x^3 y^3 + x^2 y^4 + x y^5 + x^5 z + 2 x^4 y z + 
 2 x^3 y^2 z + 2 x^2 y^3 z + 2 x y^4 z + y^5 z + x^4 z^2 + 
 2 x^3 y z^2 + 2 x^2 y^2 z^2 + 2 x y^3 z^2 + y^4 z^2 + x^3 z^3 + 
 2 x^2 y z^3 + 2 x y^2 z^3 + y^3 z^3 + x^2 z^4 + 2 x y z^4 + y^2 z^4 +
  x z^5 + y z^5$$
and
$$s_{42}(x,y,z) = (x^2 + x y + y^2) (x^2 + x z + z^2) (y^2 + y z + z^2).$$
Their dimensions are $35$ and $27$. 
A: As David Speyer also notes, this question is fairly elementary and computational; so it might better be asked first on Stack Exchange.   Aside from that, it's useful to realize that the irreducible representations of $\mathrm{SU}(3)$ are essentially the same as those of $\mathrm{SL}_3(\mathbb{C})$ or its Lie algebra.  They can be parametrized either by highest weights (Cartan-Weyl) or by partitions, and their tensor products are well-studied.   In fact, there may still be some handy online programs for computing them, as in the older Dutch project LiE.
More important, you need to distinguish clearly between a representation and its (usually different) dual even though their dimensions are the same.   In weight notation, you are considering the representations of highest weight $(2,0)$ or $(0,2)$ (of dimension 6) and those of highest weight $(2,1)$ or $(1,2)$ (of dimension 15).   But it matters which pair you actually tensor.  
ADDED: To make the result explicit in the weight notation, the easiest method follows Brauer/Racah, and gives $$(2,0) \otimes (2,1) = (4,1) \oplus (2,2) \oplus (3,0) \oplus (0,3) \oplus (1,1).$$   Weyl's dimension formula gives $\dim (a,b) = (a+1)(b+1)(a+b+2)/2$, agreeing with $90 = 35 + 27 + 10 +10 +8$.  Note that this tensor product decomposition only involves nultiplicity 1, reflecting the fact that $(2,0)$ has 6 weights with multiplicity 1.  (In the Brauer/Racah algorithm, you get a sixth weight with a coordinate $-1$ and discard it.  But in general, there are lots of cancellations and multiplicities.)
