The following comes from Exercise 13.17 of Fulton and Harris's book, Representation Theory: A First Course.
Let $V$ denote the standard representation of $\mathfrak{sl}_3\mathbb{C}$, with weights $L_1,L_2$, and $L_3$.
I would like to show that the only symmetric powers $\operatorname{Sym}^n(\operatorname{Sym}^2V)$ that contains trivial summands (e.g. $\mathbb{C}$ appears when decomposed into irreducible representations) are when $n=3k$ ($n$ is divisible by $3$), and that the trivial summand here is just the $k$-th power of the original trivial summand. So, for example, $$\operatorname{Sym}^3(\operatorname{Sym}^2V)=\operatorname{Sym}^6V\oplus \Gamma_{2,2}\oplus \mathbb{C},$$ where $\Gamma_{2,2}$ is the irreducible rep. with highest weight $2L_1-2L_3$. The goal is to say something about the appearance of $\mathbb{C}$'s for $n=3k$.
I have been hitting my head against the wall trying to do this with weight diagrams (finding $n$-wise sums of the weights of $\operatorname{Sym}^2V$ given by $\{L_i+L_j,2L_i\}$), and am having trouble finding a general pattern to the decomposition of $\operatorname{Sym}^n(\operatorname{Sym}^2V)$. I also know that the dimension of $\operatorname{Sym}^n(\operatorname{Sym}^2V)$ is ${n+5}\choose n$ and the dimension of each irreducible rep. with highest weight $aL_1-bL_3$ is $\dim(\Gamma_{a,b})=(a+1)(b+1)(\frac{a+b+2}{2})$, but I do not think this helps much.
If you look at this from an algebraic geometric standpoint, apparently the existence of this trivial summand in the decomposition above tells us that there exists a cubic hypersurface $X$ in $\mathbb{P}^5$ preserved under all automorphisms of $\mathbb{P}^5$ carrying the Veronese surface to itself. But my background here is lacking, and this is all Greek to me.
If anyone could possibly recommend a better way to approach this problem, I would greatly, greatly appreciate it.