3. Given that $a, b$ are the real roots of the equation $x^{2}+2 p x+3=0$, $c, d$ are the real roots of the equation $x^{2}+2 q x+3=0$, and $e, v$ are the real roots of the equation $x^{2}+2\left(p^{2}-q^{2}\right) x+3=0$, where $p, q$ are real numbers. Then, the value of $\frac{(a-c)(b-c)(a+d)(b+d)}{e+v}$ is $\qquad$ .
3.6 .

From the relationship between roots and coefficients, we get
$$
\begin{array}{l}
a+b=-2 p, a b=3 ; c+d=-2 q, c d=3 ; \\
e+f=-2\left(p^{2}-q^{2}\right), e f=3 . \\
\text { Therefore } \frac{(a-c)(b-c)(a+d)(b+d)}{e+f} \\
=\frac{\left[a b-(a+b) c+c^{2}\right]\left[a b+(a+b) d+d^{2}\right]}{e+f} \\
=\frac{\left(c d+2 p c+c^{2}\right)\left(c d-2 p d+d^{2}\right)}{-2\left(p^{2}-q^{2}\right)} \\
=\frac{c d(d+2 p+c)(c-2 p+d)}{-2\left(p^{2}-q^{2}\right)} \\
=\frac{3\left[(c+d)^{2}-4 p^{2}\right]}{-2\left(p^{2}-q^{2}\right)}=\frac{3\left(4 q^{2}-4 p^{2}\right)}{-2\left(p^{2}-q^{2}\right)}=6 .
\end{array}
$$