183 If $a, b, c \in [0,1]$, then the maximum value of the ternary function
$$
f(a, b, c)=a(1-a+a \cdot b)(1-a b+a b c)(1-c)
$$

is =
$183 \frac{8}{27}$. Clearly, $f(a, b, c)=(1-a+a b)(1-a b+a b c)(a-a c)$, and
$$
1-a+a b \geqslant 0,1-a b+a b c \geqslant 0, a-a c \geqslant 0 .
$$

By the AM-GM inequality, we have
$$
\begin{aligned}
f(a, b, c) & \leqslant\left(\frac{1-a+a b+1-a b+a b c+a-a c}{3}\right)^{3} \\
& =\left(\frac{2+a b c-a c}{3}\right)^{3} \leqslant\left(\frac{2}{3}\right)^{3}=\frac{8}{27},
\end{aligned}
$$

Moreover, when $a=\frac{2}{3}, b=\frac{1}{2}, c=0$, the equality holds, hence the maximum value of $f(a, b, c)$ is $\frac{8}{27}$.