In Kerson Huang's Quantum Field Theory 2nd, p 278, the author introduces "cumulant expansion" from
$$ \ln \langle e^x \rangle = \ln \sum_{n=0}^\infty \frac{\langle x^n \rangle}{n!} = \langle x \rangle + \frac{1}{2} [ \langle x^2 \rangle - \langle x \rangle^2 ] + \frac{1}{6} [ \langle x^3 \rangle - 3 \langle x \rangle \langle x^2 \rangle + \langle x\rangle^3 ] + \cdots (14.112) $$
However, if I start from $$\ln \langle e^x \rangle = \ln( 1 + \langle x \rangle + \frac{\langle x^2 \rangle}{2!} + \frac{ \langle x^3 \rangle}{3!} + \cdots) $$
use the Taylor expansion $\ln (1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} + \cdots$ I got $$ \langle x \rangle + \frac{\langle x^2\rangle}{2!} +\frac{ \langle x^3 \rangle}{3!} + \cdots - \frac{1}{2} \left(\langle x \rangle + \frac{\langle x^2 \rangle }{2!}+ \cdots \right)^2 + \frac{1}{3} \left( \langle x \rangle + \cdots \right)^3 $$
I have the terms at $x$ and $x^2$ correct. For the cubic, I have $$ \frac{1}{6} \left( \langle x^3 \rangle - 3 \langle x\rangle \langle x^2 \rangle + 2 \langle x \rangle ^3 \right)$$ Why the $\langle x \rangle^3$ differs with the prefactor?
