6. Given $x, y, z \in R$, and $x+y+z=2, x^{2}+y^{2}+z^{2}=14, x^{3}+y^{3}+z^{3}=17$, then $x y z=$
6. -7

Explanation: From $x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z) \cdot\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$
$$
\begin{array}{l}
\Leftrightarrow x y z=\frac{x^{3}+y^{3}+z^{3}-(x+y+z) \cdot\left(x^{2}+y^{2}+z^{2}-\frac{(x+y+z)^{2}-\left(x^{2}+y^{2}+z^{2}\right)}{2}\right)}{3} \\
\text { Substituting the conditions, we get: } x y z=\frac{17-2 \cdot\left(14-\frac{4-14}{2}\right)}{3}=-7
\end{array}
$$