$$
\begin{array}{l}
\cos 10^{\circ} \cdot \cos 50^{\circ} \cdot \cos 70^{\circ}+\sin 10^{\circ} \cdot \sin 50^{\circ} \cdot \sin 70^{\circ} \\
=\quad .
\end{array}
$$
6. $\frac{1+\sqrt{3}}{8}$.

Original expression $=\sin 20^{\circ} \cdot \sin 40^{\circ} \cdot \sin 80^{\circ}+\cos 20^{\circ} \cdot \cos 40^{\circ} \cdot \cos 80^{\circ}$ is easy to calculate
$$
\begin{array}{l}
8 \sin 20^{\circ} \cdot \sin 40^{\circ} \cdot \sin 80^{\circ} \\
=4\left(\cos 20^{\circ}-\cos 60^{\circ}\right) \sin 80^{\circ} \\
=4 \sin 80^{\circ} \cdot \cos 20^{\circ}-2 \sin 80^{\circ} \\
=2\left(\sin 100^{\circ}+\sin 60^{\circ}\right)-2 \sin 80^{\circ} \\
=2 \sin 60^{\circ}=\sqrt{3}, \\
8 \cos 20^{\circ} \cdot \cos 40^{\circ} \cdot \cos 80^{\circ} \\
=4\left(\cos 60^{\circ}+\cos 20^{\circ}\right) \cos 80^{\circ} \\
=2 \cos 80^{\circ}+4 \cos 20^{\circ} \cdot \cos 80^{\circ} \\
=2 \cos 80^{\circ}+2\left(\cos 100^{\circ}+\cos 60^{\circ}\right) \\
=2 \cos 80^{\circ}-2 \cos 80^{\circ}+2 \cos 60^{\circ} \\
=1 .
\end{array}
$$

Thus, the original expression $=\frac{1+\sqrt{3}}{8}$.