6. $a, b, c$ are all positive integers, and satisfy $a b+b c=$ $3984, a c+b c=1993$. Then the maximum value of $a b c$ is $\qquad$

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6. Since 1993 is a prime number, from $c(a+b)=1993$, we get $c=1, a+b=1993$. Substituting $b=1993-a$ into the other equation, we get $a^{2}-1992 a+1991=0$. Solving this, we get $a_{1}=1, a_{2}=1991$, then $b_{1}=1992, b_{2}=2$. Therefore, $(a, b, c)$ has two sets $(1,1992,1),(1991,2,1)$. Thus, the maximum value of $abc$ is 3982.