Answer: -8
Solution: Since $a, b, c$ are distinct and $a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)=0$, then $a+b+c=0$. From $a+b+c=0$ and $a^{2}+b^{2}+c^{2}=8$, we have $a b+b c+c a=-4$ and $a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}=(a b+b c+c a)^{2}-2 a b c(a+b+c)=(a b+b c+c a)^{2}=16$. Thus,
$$
\frac{a b}{a+b}+\frac{b c}{b+c}+\frac{c a}{c+a}=\frac{a b}{-c}+\frac{b c}{-a}+\frac{c a}{-b}=\frac{-\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right)}{a b c}=-\frac{16}{2}=-8
$$
10. If $a^{3}+b^{3}+c^{3}=3 a b c=6$ and $a^{2}+b^{2}+c^{2}=8$, find the value of $\frac{a b}{a+b}+\frac{b c}{b+c}+\frac{c a}{c+a}$.