My understanding is from using both Humphreys and Fulton-Harris.
Usually, given a Lie algebra (in this case, I want to take $\mathfrak{g} = \mathfrak{sl}_3 (\mathbb{C})$) and an irreducible representation $V$, we decompose $V$ into its weight spaces with respect to some Cartan (again in this case, $\mathfrak{h}$ is the diagonal traceless matrices), then find a highest weight vector killed by all positive weight spaces (fixing an ordering on roots beforehand), and this tells us the weight of $V$.
In the other direction, given some weight $\lambda$, the usual construction is to build a (huge) highest weight module $Z(\lambda)$ and quotient this out by its unique maximal submodule to get the desired representation $V(\lambda)$. In practice however, this seems more complicated to me than it could be.
The specific example I want to work with is using the weight $\lambda = (1,0,-1)$ in the weight space with basis given by $L_i$ matrices which send a diagonal matrix to its $i$-th entry. The corresponding representation is the adjoint rep, $\mathfrak{sl}_3$ acting on itself.
I know from results given in e.g. Fulton-Harris, the representations are contained in a tensors of symmetric powers of the various wedge products of the standard representation $V = \mathbb{C}^3$ - however, I can't see how $\mathfrak{sl}_3$ lives in this tensor product.
So, my question is:
Is there a simpler way of seeing that the weight $(1,0,-1)$ corresponds to $V = \mathfrak{sl}_3$? What about other weights, such as $(3,0,0)$ which corresponds to the symmetric algebra, or $(1,1,1)$?
The only way I see this to work is to 'guess' a highest weight vector. In the $(1,0,-1)$ case, I can 'guess' that $V = \mathfrak{sl}_3$, and then find the highest weight vector to be a matrix with 1 in the top right entry. Same as in $(3,0,0)$, I can 'guess' the highest weight monomial. But this won't work for other examples, so I'm looking for something a bit more concrete.