variational calculus - BRST Quantization of the Bosonic String: Nilpotence of BRST transformation (Polchinski)

Currently I am studying string theory and I encountered a bunch of interrelated problems in the context of BRST quantization which I can't solve for myself although I tried hard for some days.

My question concerns the BRST transformation of the bosonic string in eq. (4.3.1) of Polchinski's book. In the paragraph following these transformations, Polchinski says that "the reader can check nilpotence up to the equations of motion". I tried but wasn't able to complete the proof - here my calculations:

$$\begin{align} \delta_\mathrm{B}\delta_\mathrm{B}'X^\mu&=\delta_\mathrm{B}\left[\mathrm{i}\varepsilon'\left(c\partial X^\mu+\bar{c}\bar{\partial}X^\mu\right)\right]\\[2pt] &=\mathrm{i}\varepsilon'\left[(\delta_\mathrm{B}c)\partial X^\mu+c(\delta_\mathrm{B}\partial X^\mu)+(\delta_\mathrm{B}\bar{c})\partial X^\mu+\bar{c}(\delta_\mathrm{B}\bar{\partial} X^\mu)\right]\\[2pt] &=\mathrm{i}\varepsilon'\left[(\mathrm{i}\varepsilon c\partial c)\partial X^\mu+c(\mathrm{i}\varepsilon(c\partial\partial X^\mu+\bar{c}\underbrace{\bar{\partial}\partial X^\mu}_{=0}))+(\mathrm{i}\varepsilon\bar{c}\bar{\partial}\bar{c})\bar{\partial}X^\mu+\bar{c}(\mathrm{i}\varepsilon(c\underbrace{\partial\bar{\partial}X^\mu}_{=0}+\bar{c}\bar{\partial}\bar{\partial}X^\mu))\right]\\[2pt] &=-\varepsilon'\left[\varepsilon c\partial c\partial X^\mu+c\varepsilon c\partial\partial X^\mu+\varepsilon\bar{c}\bar{\partial}\bar{c}\bar{\partial}X^\mu+\bar{c}\varepsilon\bar{c}\bar{\partial}\bar{\partial}X^\mu\right]\\[2pt] &=-\varepsilon'\varepsilon \left[c\partial c\partial X^\mu-\underbrace{c c}_{=0}\partial\partial X^\mu+\bar{c}\bar{\partial}\bar{c}\bar{\partial}X^\mu-\underbrace{\bar{c}\bar{c}}_{=0}\bar{\partial}\bar{\partial}X^\mu\right]\\[2pt] &=-\varepsilon'\varepsilon \left[c\partial c\partial X^\mu+\bar{c}\bar{\partial}\bar{c}\bar{\partial}X^\mu\right]\\[2pt] &=? \end{align}$$

Here I used the equation of motion for the field $X^\mu$ and exploited $c^2=0=\bar{c}^2$. At this point, however, I do not see why the remaining terms should vanish.

Answer

Hint: The infinitesimal BRST transformation

$$\delta_{\mathrm{B}}\left(F[X,c,\ldots]\right)~:=~F[X+\delta_{\mathrm{B}}X,c+\delta_{\mathrm{B}}c,\ldots]-F[X,c,\ldots]\tag{A}$$ acts "under" the spacetime derivatives marked in red, i.e. OP's second line should read $$\begin{align} \delta_{\mathrm{B}}\delta_\mathrm{B}'X^\mu &\stackrel{(4.3.1a)}=\delta_\mathrm{B}\left[\mathrm{i}\varepsilon'\left(c\color{red}{\partial} X^\mu+\bar{c}\color{red}{\bar{\partial}}X^\mu\right)\right]\\[2pt] &~~~\stackrel{(A)}=~~~\mathrm{i}\varepsilon'\left[(c+\delta_{\mathrm{B}}c)\color{red}{\partial} (X^\mu+\delta_{\mathrm{B}}X^\mu)+(\bar{c}+\delta_{\mathrm{B}}\bar{c})\color{red}{\bar{\partial}}(X^\mu+\delta_{\mathrm{B}}X^\mu)-c\color{red}{\partial} X^\mu-\bar{c}\color{red}{\bar{\partial}}X^\mu\right]\\[2pt] &~~~=~~~\mathrm{i}\varepsilon'\left[(\delta_\mathrm{B}c)\color{red}{\partial} X^\mu +c\color{red}{\partial}(\delta_\mathrm{B}X^\mu) +(\delta_\mathrm{B}\bar{c})\color{red}{\bar{\partial}} X^\mu +\bar{c}\color{red}{\bar{\partial}}(\delta_\mathrm{B}X^\mu)\right]\\[2pt] &~~~=~~~\ldots\tag{B} \end{align}$$