The quadratic Casimir eigenvalue on a highest-weight module is $(\lambda ,\lambda +2\rho )$

By Central elements act by scalars on cyclic highest-weight modules and The quadratic Casimir element is central, it is enough to compute $Cv$ on the highest vector $v$. The root-system theorem The root set is a reduced crystallographic root system makes every root space one-dimensional, while nondegeneracy and root-space orthogonality make ${\mathfrak{g}}_{\alpha }$ and ${\mathfrak{g}}_{-\alpha }$ dual. Choose a basis of $\mathfrak{h}$ and root vectors $e_{\alpha }\in {\mathfrak{g}}_{\alpha }$, $f_{\alpha }\in {\mathfrak{g}}_{-\alpha }$ with $B(e_{\alpha },f_{\alpha })=1$; these are full dual bases, so the Casimir decomposes as $C={\sum }_j{h}_j{h}^j+{\sum }_{\alpha >0}(e_{\alpha }{f}_{\alpha }+f_{\alpha }{e}_{\alpha })$.

Since $e_{\alpha }v=0$ for every positive root, one has $f_{\alpha }{e}_{\alpha }v=0$ and $e_{\alpha }{f}_{\alpha }v=[e_{\alpha },f_{\alpha }]v$. By Opposite root spaces bracket to the Killing-dual line, that bracket is $H_{\alpha }$, so step 1.1 gives $Cv=({\sum }_j{h}_j{h}^j+{\sum }_{\alpha >0}{H}_{\alpha })v$.

The Cartan part acts on $v$ by $(\lambda ,\lambda )$, and the root contribution acts by ${\sum }_{\alpha >0}\lambda (H_{\alpha })=2(\lambda ,\rho )$. Therefore $Cv=(\lambda ,\lambda +2\rho )v$, so the Casimir scalar on $M$ is $(\lambda ,\lambda +2\rho )$.