6. If three angles $x, y, z$ form an arithmetic sequence with a common difference of $\frac{\pi}{3}$, then $\tan x \cdot \tan y + \tan y \cdot \tan z + \tan z \cdot \tan x$ $=$ . $\qquad$
6. -3 .
From the problem, we know that
$1+\tan x \cdot \tan y=\frac{\tan y-\tan x}{\tan (y-x)}=\frac{\tan y-\tan x}{\sqrt{3}}$.
Similarly, $1+\tan y \cdot \tan z=\frac{\tan z-\tan y}{\sqrt{3}}$,
$1+\tan z \cdot \tan x=\frac{\tan z-\tan x}{-\sqrt{3}}$.
Adding the above three equations, we get
$\tan x \cdot \tan y+\tan y \cdot \tan z+\tan z \cdot \tan x=-3$.