10. The function $y=f(x)$ defined on $\mathbf{R}$ has the following properties:
（1）For any $x \in \mathbf{R}$, $f\left(x^{3}\right)=f^{3}(x)$;
(2) For any $x_{1} 、 x_{2} \in \mathbf{R}, x_{1} \neq x_{2}$, $f\left(x_{1}\right)$ $\neq f\left(x_{2}\right)$.
Then the value of $f(0)+f(1)+f(-1)$ is $\qquad$
10.0 .

From $f(0)=f^{3}(0)$, we know $f(0)[1-f(0)][1+f(0)]=0$,
thus, $f(0)=0$ or $f(0)=1$, or $f(0)=-1$;
from $f(1)=f^{3}(1)$, similarly $f(1)=0$ or 1 or 1;
from $f(-1)=f^{3}(-1)$, similarly $f(-1)=0$ or 1 or -1.
However, $f(0)$, $f(1)$, and $f(-1)$ are pairwise distinct,
so $\{f(0), f(1), f(-1)\}=\{0,1,-1\}$.
Therefore, $f(0)+f(1)+f(-1)=0$.