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The Faddeev-Popov action reads $$S_3=b_Ac^{\alpha}\delta_{\alpha}F^A(\phi).\tag{4.2.5}$$ I want to find the BRST variation of the gauge variation of $F^A$ in $S_3$ i.e. $$b_Ac^{\alpha}\color{red}{\delta_{B}(}\delta_{\alpha}F^A(\phi)\color{red}{)}.\tag{*}$$ I did following manipulations $$b_Ac^{\alpha}(f^{\gamma}_{B\alpha}\delta_{\gamma}+\delta_{\alpha}\delta_{B})F^A(\phi)$$ $$=b_Ac^{\alpha}\Big(f^{\gamma}_{B\alpha}\delta_{\gamma}F^A(\phi)+\delta_{\alpha}\delta_{B}F^A(\phi)\Big)$$ $$=b_Ac^{\alpha}\Big(f^{\gamma}_{B\alpha}\delta_{\gamma}F^A+\delta_{\alpha}(-i\epsilon c^\gamma\delta_{\gamma}F^A)\Big)$$ $$=b_Ac^{\alpha}\Big(f^{\gamma}_{B\alpha}\delta_{\gamma}F^A-i\epsilon\delta_{\alpha}( c^\gamma)\delta_{\gamma}F^A-\color{green}{i\epsilon c^\gamma\delta_{\alpha}\delta_{\gamma}F^A}\Big).$$ I can continue this manipulation for the green colored termed. There are 2 strange things that happened with this calculation

  1. Variation produce finite term ($1$st term in last expression) instead of infinitesimal quantity
  2. This variation doesn't cancel out of the variation of $c^{\alpha}$ in $S_3$ which is $0$ shown here.

Where am I making the mistake? We have $\delta_{\alpha}$ in $S_3$ because of expansion of $\delta(F^a({\phi}))$ around gauge condition $F^A(\phi)=0$.

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OP's eq. (*) should read $$\begin{align} b_Ac^{\alpha}\delta_B\delta_{\alpha}F^A(\phi)~=~~&b_Ac^{\alpha}(\delta_B\phi^i)\frac{\delta}{\delta \phi^i}\delta_{\alpha}F^A(\phi)\cr ~\stackrel{(4.2.6a)}{=}&b_Ac^{\alpha}(-i\epsilon c^{\beta})(\delta_{\beta}\phi^i)\frac{\delta}{\delta \phi^i}\delta_{\alpha}F^A(\phi)\cr ~=~~&-i\epsilon b_Ac^{\alpha}c^{\beta}\delta_{\beta}\delta_{\alpha}F^A(\phi)\cr ~=~~&-\frac{i\epsilon}{2} b_Ac^{\alpha}c^{\beta}[\delta_{\beta},\delta_{\alpha}]F^A(\phi)\cr ~\stackrel{(4.2.1)}{=}&-\frac{i\epsilon}{2} b_Ac^{\alpha}c^{\beta}f^{\gamma}_{\beta\alpha}\delta_{\gamma}F^A(\phi). \end{align} $$

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