6. Given $p(x)=a x^{3}+b x^{2}+c x+d$ is a cubic polynomial, satisfying $p\left(\frac{1}{2}\right)+p\left(-\frac{1}{2}\right)=1000 p(0)$. Let $x_{1}, x_{2}$, $x_{3}$ be the 3 roots of $p(x)=0$, then the value of $\frac{1}{x_{1} x_{2}}+\frac{1}{x_{2} x_{3}}+\frac{1}{x_{1} x_{3}}$ is $\qquad$.
Solve
$$
\left(\frac{1}{8} a+\frac{1}{4} b+\frac{1}{2} c+d\right)+\left(-\frac{1}{8} a+\frac{1}{4} b-\frac{1}{2} c+d\right)=1000 d \Rightarrow \frac{1}{2} b=998 d \text {. }
$$

By Vieta's formulas, $\left\{\begin{array}{l}x_{1}+x_{2}+x_{3}=-\frac{b}{a}, \\ x_{1} x_{2} x_{3}=-\frac{d}{a}\end{array} \Rightarrow \frac{x_{1}+x_{2}+x_{3}}{x_{1} x_{2} x_{3}}=\frac{b}{d}=1996\right.$,
so $\frac{1}{x_{1} x_{2}}+\frac{1}{x_{2} x_{3}}+\frac{1}{x_{1} x_{3}}=\frac{x_{1}+x_{2}+x_{3}}{x_{1} x_{2} x_{3}}=1996$.