1.64 If $a<b$, then the radical $\sqrt{-(x+a)^{3}(x+b)}$ is equal to
(A) $(x+a) \sqrt{-(x+a)(x+b)}$.
(B) $(x+a) \sqrt{(x+a)(x+b)}$.
(C) $-(x+a) \sqrt{-(x+a)(x+b)}$.
(D) $-(x+a) \sqrt{(x+a)(x+b)}$.
(China Junior High School Mathematics League, 1986)
[Solution] $\because \quad-(x+a)^{3}(x+b) \geqslant 0$,
that is
$$
(x+a)(x+b) \leqslant 0 \text {, }
$$

and $\because a<b$, then $-b \leqslant x \leqslant-a$.
$$
\therefore \quad \text { the original expression }=-(x+a) \sqrt{-(x+a)(x+b)} \text {. }
$$

Therefore, the answer is $(C)$.