8. Given $x, y \in[0,+\infty)$, then the minimum value of $x^{3}+y^{3}-5 x y$ is $\qquad$ .
Let $x=y \Rightarrow u=2 x^{3}-5 x^{2}, u^{\prime}=6 x^{2}-10 x=2 x(3 x-5)$, we can see that the original expression achieves its minimum value at $x=y=\frac{5}{3}$. Therefore, $x^{3}+y^{3}-5 x y=x^{3}+y^{3}+\left(\frac{5}{3}\right)^{3}-5 x y-\left(\frac{5}{3}\right)^{3} \geqslant 3 x y \cdot \frac{5}{3}-5 x y-\frac{125}{27}=-\frac{125}{27}$.