I'm trying to figure out the form of the Casimir operators for $su(3)$. According to Racah's theorem, this group has two independent Casimir operators that can be constructed from its generators. The first one is easy to find: using the Gell-Mann matrices $\lambda_i,\ i=1,...8\ $ as generators, it can be written as
$$ C_1 = \sum_{i=1}^8 \lambda_i^2 $$
However, I'm struggling to construct the second one. In $\textit{The Lie Algebras su(N). An introduction}$ (section 4.11), Walter Pfeifer writes that $C_2$ is given by $C_2 = C_1(2C_1 - 11/6)$, and that this operator can be transformed so that it is expressed as sum of products of three $\lambda_i$'s. However, this Casimir operator is not independent from $C_1$ (in fact, it is completely determined by it). So, is this expression correct? If not, how can one construct the missing Casimir operator for $su(3)$?
I know for a fact that the form of the second Casimir of $SU(3)$ as constructed from the Gell-Mann Matrices is:
$$ d^{abc}T^aT^bT^c $$
Where $T^a = \frac{\lambda^a}{2}$ (Sorry but as a physicist this is the basis I am the most familiar with) and $d^{abc}$ are the symmetric symbols, defined as $2 \, Tr(\{T^a,T^b\}T^c)$.
Sadly I don't remember the proof that this product actually commutes with all the $T^d$ and is therefore a Casimir. I tried doing it before writing this answer but couldn't make it work. I hope that my answer revitalizes this question and someone can direct the both of us to a proof.
P.S. I used Einstein's summation notation in my answer, i later saw that you didn't, i hope this notation isn't unfamiliar to you, just assume that avery repeated index is summed.