$7 \cdot 103$ A function $y=f(x)$ defined on $R$ has the following properties:
(1) For any $x \in R$, there is $f\left(x^{3}\right)=f^{3}(x)$;
(2) For any $x_{1}, x_{2} \in R, \quad x_{1} \neq x_{2}$, there is $f\left(x_{1}\right) \neq f\left(x_{2}\right)$. Then the value of $f(0)+f(1)+f(-1)$ is
(A) 0.
(B) 1.
(C) -1.
(D) Uncertain.
(2nd “Hope Cup” National Mathematics Invitational Competition, 1991)
[Solution] By the given condition, we notice that $f(0)=f^{3}(0)$, which implies
$$
f(0)[f(0)-1][f(0)+1]=0 .
$$

Thus,
$$
f(0)=0,1 \text {, or }-1 \text {. }
$$

Also, $f(1)=f^{3}(1)$, similarly we know $f(1)=0,1$, or -1.
Furthermore, by $f(-1)=f^{3}(-1)$, we also know $f(-1)=0,1$, or -1. However, $f(-1), f(0), f(1)$ are all distinct, hence
$$
\{f(-1), f(0), f(1)\}=\{-1,0,1\} \text {. }
$$

Therefore,
$$
f(-1)+f(0)+f(1)=0 \text {. }
$$

Thus, the answer is $(A)$.