6. Let $a, b$ be the roots of the equation $x^{2}+68 x+1=0$, and $c, d$ be the roots of the equation $x^{2}-86 x+1=0$. Then
$$
(a+c)(b+c)(a-d)(b-d)
$$

the value is $\qquad$.
6. 2772 .
$$
\begin{array}{l}
\text { Given } a b=c d=1, a+b=-68, \\
c^{2}-8 b c+1=d^{2}-86 d+1=0, \\
\text { then }(a+c)(b+c)(a-d)(b-d) \\
=\left[a b+(a+b) c+c^{2}\right]\left[a b-(a+b) d+d^{2}\right] \\
=\left(1-68 c+c^{2}\right)\left(1+68 d+d^{2}\right) \\
=18 c \cdot 154 d=2772 .
\end{array}
$$