8. Given real numbers $a, b, x, y$ satisfy $a+b=x+y=2$, $a x+b y=5$. Then the value of $\left(a^{2}+b^{2}\right) x y+a b\left(x^{2}+y^{2}\right)$ is $\qquad$ .
8. -5 .

Given $a+b=x+y=2$, we have
$$
(a+b)(x+y)=a x+b y+a y+b x=4 \text{. }
$$

Since $a x+b y=5$, then,
$$
\begin{array}{l}
a y+b x=-1 . \\
\text{ Thus, }\left(a^{2}+b^{2}\right) x y+a b\left(x^{2}+y^{2}\right) \\
=(a y+b x)(a x+b y)=-5 .
\end{array}
$$