8. Given $x, y \in [0,+\infty)$. Then the minimum value of $x^{3}+y^{3}-5 x y$ is $\qquad$ .
8. $-\frac{125}{27}$.

Let $x+y=t$.
Since $x^{3}+y^{3} \geqslant x^{2} y+x y^{2}$, we have
$x^{3}+y^{3} \geqslant \frac{1}{4}(x+y)^{3}$.
Also, $(x+y)^{2} \geqslant 4 x y$, so
$x^{3}+y^{3}-5 x y \geqslant \frac{1}{4}(x+y)^{3}-\frac{5}{4}(x+y)^{2}$
$=\frac{1}{4}\left(t^{3}-5 t^{2}\right)$.
Let $f(t)=t^{3}-5 t^{2}$, then
$f^{\prime}(t)=3 t^{2}-10 t$.
Setting $f^{\prime}(t)=0$, we find $t=0$ or $\frac{10}{3}$.
Since $x, y \in[0,+\infty)$, it is not difficult to analyze that when $t=\frac{10}{3}$, $f(t)$ reaches its minimum value $-\frac{125}{27}$.
Therefore, when $x=y=\frac{5}{3}$, the minimum value of the original expression is $-\frac{125}{27}$.