8. Given real numbers $a, b$ satisfy
$$
a^{2}+a b+b^{2}=1 \text {, and } t=a b-a^{2}-b^{2} \text {. }
$$

Then the product of the maximum and minimum values of $t$ is $\qquad$
8. 1.

Let $a=x+y, b=x-y$. Then
$$
(x+y)^{2}+(x+y)(x-y)+(x-y)^{2}=1 \text {. }
$$

Simplifying, we get $y^{2}=1-3 x^{2}$.
Since $y^{2} \geqslant 0$, we have $0 \leqslant x^{2} \leqslant \frac{1}{3}$.
Thus, $t=a b-a^{2}-b^{2}$
$$
\begin{array}{l}
=(x+y)(x-y)-(x+y)^{2}-(x-y)^{2} \\
=-x^{2}-3 y^{2}=8 x^{2}-3 .
\end{array}
$$

Therefore, $-3 \leqslant t \leqslant-\frac{1}{3}$.
Hence, $-3 \times\left(-\frac{1}{3}\right)=1$.