= Solution
The commuting <Hermitian operators> $H_1,H_2$ admit a <weight-space decomposition>. The six <root vectors> shift their simultaneous eigenvalues by
$$
\alpha_1=(1,0),\qquad\alpha_2=(-1/2,\sqrt3/2),\qquad\alpha_3=(1/2,\sqrt3/2)
$$
or their negatives. Put $h_i=2\alpha_i\cdot H$, $e_i=\sqrt2E_+^i$ and $f_i=\sqrt2E_-^i$. The supplied relations give
$$
[h_i,e_i]=2e_i,\qquad[h_i,f_i]=-2f_i,\qquad[e_i,f_i]=h_i.
$$
These are three <sl2 triples>, coming from the complexifications of the three <SU(2) Lie algebra> root-pair subalgebras. They act through homomorphisms into $\operatorname{End}(V)$. For a nontrivial representation, simplicity of the <SU(3) Lie algebra> makes its kernel zero, so these images are actual $\mathfrak{sl}_2(\mathbb C)$ subalgebras; a trivial representation gives zero actions. The real compact generators can be taken as $ih_i$, $i(E_+^i+E_-^i)$ and $E_+^i-E_-^i$.
Decomposing under any one of these subalgebras gives finite <weight strings> in its root direction. The two simple-root highest weights must be nonnegative integers. In the <SU(3) highest-weight coordinates> used here, this condition is
$$
\boxed{a=2p\in\mathbb Z_{\geq0},\qquad b=\sqrt3q-p\in\mathbb Z_{\geq0}.}
$$
Here $(a,b)$ are the <Dynkin labels>; $(p,q)$ are the eigenvalues of $H_1,H_2$. An <irreducible representation> is determined by these labels and has dimension $(a+1)(b+1)(a+b+2)/2$. An arbitrary finite-dimensional anti-Hermitian representation is a direct sum of such representations: the <orthogonal complement> of an <invariant subspace> remains invariant.
The <weight diagram> has <Weyl group> symmetry, namely reflections and rotations by $120$ degrees. Its convex boundary is a hexagon with alternating side lengths $a,b$, degenerating to a triangle when one label vanishes. Generic representations need not have $60$-degree symmetry. Boundary <weight multiplicities> are one. Moving inward through hexagonal layers increases multiplicity by one, up to $\min(a,b)+1$; triangular inner layers retain that maximum. An exact version of this rule is the <dominant weight multiplicity formula for sl3>: move a weight by the Weyl group to dominant Dynkin coordinates $(c,d)$ and set
$$
r=\frac{2a+b-2c-d}{3},\qquad s=\frac{a+2b-c-2d}{3}.
$$
Its multiplicity is zero unless $r,s$ are integers, and otherwise is $\max\{0,1+\min(a,b,r,s)\}$. These rules construct the diagram while counting coincident weights correctly.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/iii/paper-49-su3-weights.png]
{title=SU(3) root directions and weight diagrams, with multiplicities at coincident weights}
For the Casimir calculation, set $t_3=H_1$, $t_8=H_2$, and for each root pair use $(E_++E_-)/\sqrt2$ and $(E_+-E_-)/(i\sqrt2)$. In the defining representation these eight generators satisfy $\operatorname{tr}(t_at_b)=\delta_{ab}/2$. Their summed squares are precisely the question's $X$. The invariant trace form makes their <structure constants of a Lie algebra> antisymmetric, so
$$
[X,t_c]=\sum_a\{t_a,[t_a,t_c]\}=0:
$$
the antisymmetric coefficient contracts a symmetric anticommutator. Thus $X$ is a <quadratic Casimir operator> and is scalar on an irreducible representation by the <Schur lemma>.
Let $v$ be a <highest-weight vector>, annihilated by all $E_+^i$. On $v$, each $E_-^iE_+^i$ vanishes and each $E_+^iE_-^i$ reduces to the root commutator. Summing those three commutators gives $H_1+\sqrt3H_2$. Hence
$$
Xv=(p^2+q^2+p+\sqrt3q)v,
$$
and therefore the <SU(3) quadratic Casimir eigenvalue> is
$$
\boxed{X=(p+\sqrt3q+p^2+q^2)I_V
=\frac{a^2+ab+b^2+3a+3b}{3}I_V.}
$$
For a reducible representation the same calculation applies separately to its irreducible blocks.
For light-hadron <flavour symmetry>, the <up quark>, <down quark> and <strange quark> form $\mathbf3$, and <antiquarks> form $\overline{\mathbf3}$. Conventional <mesons> have $\mathbf3\otimes\overline{\mathbf3}=\mathbf1\oplus\mathbf8$. Conventional three-quark <baryons> have
$$
\mathbf3^{\otimes3}=\mathbf1\oplus\mathbf8\oplus\mathbf8\oplus\mathbf{10},
$$
obtained from $\mathbf3\otimes\mathbf3=\mathbf6\oplus\overline{\mathbf3}$, $\mathbf6\otimes\mathbf3=\mathbf{10}\oplus\mathbf8$ and $\overline{\mathbf3}\otimes\mathbf3=\mathbf8\oplus\mathbf1$. Their <Dynkin labels> are $(0,0)$, $(1,1)$ and $(3,0)$, respectively. Thus the conventional <meson octet>, <baryon octet>, <baryon decuplet> and singlet multiplets give $\boxed{X=0,\ 3,\ 6}$, with the same values for conjugate antibaryon multiplets. More generally, the <zero flavour triality of a light-quark hadron> imposes $a+2b\equiv0\pmod3$ on multiplets constructed entirely from light quarks, <antiquarks> and <gluons>. Their Casimir values are given by the same formula for these labels. Higher multiquark representations can therefore have other eigenvalues; the illustrated $\mathbf{27}$, of labels $(2,2)$, has $X=8$.
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