quantum field theory - Calculating the commutator of Pauli-Lubanski operator and generators of Lorentz group

The Pauli-Lubanski operator is defined as $${W^\alpha } = \frac{1}{2}{\varepsilon ^{\alpha \beta \mu \nu }}{P_\beta}{M_{\mu \nu }},\qquad ({\varepsilon ^{0123}} = + 1,\;{\varepsilon _{0123}} = - 1)$$ where $M_{\mu\nu}$ is the generators of Lorentz group.

The commutation relation between generators of Poincare group is know as $$i[{M^{\mu \nu }},{M^{\rho \sigma }}] = {\eta ^{\nu \rho }}{M^{\mu \sigma }} - {\eta ^{\mu \rho }}{M^{\nu \sigma }} - {\eta ^{\mu \sigma }}{M^{\rho \nu }} + {\eta ^{\nu \sigma }}{M^{\rho \mu }},$$ $$i[{P^\mu },{M^{\rho \sigma }}] = {\eta ^{\mu \rho }}{P^\sigma } - {\eta ^{\mu \sigma }}{P^\rho }.$$

I try to derive the commutator between Pauli-Lubanski operator and a generator of Lorentz group, which is also given in our lecture $$i[{W^\alpha },{M^{\rho \sigma }}] = {\eta ^{\alpha \rho }}{W^\sigma } - {\eta ^{\alpha \sigma }}{W^\rho }.\tag{*}$$

But I only get $$\begin{align} i[{W^\alpha },{M^{\rho \sigma }}] =& i[\frac{1}{2}{\varepsilon ^{\alpha \beta \mu \nu }}{P_\beta }{M_{\mu \nu }},{M^{\rho \sigma }}] \\ =& \frac{1}{2}{\varepsilon ^{\alpha \beta \mu \nu }}{P_\beta }\left( {\delta _\nu ^\rho {M_\mu }^\sigma - \delta _\mu ^\rho {M_\nu }^\sigma - \delta _\mu ^\sigma {M^\rho }_\nu + \delta _\nu ^\sigma {M^\rho }_\mu } \right) \\ & + \frac{1}{2}{\varepsilon ^{\alpha \beta \mu \nu }}\left( {\delta _\beta ^\rho {P^\sigma } - \delta _\beta ^\sigma {P^\rho }} \right)M_{\mu\nu}. \end{align}$$

Obviously I can contract out the delta's, but that does not get me any closer to the simpler result of (*). Can anyone point out what to do next?