(5) The equation $x^{3}-12 x+a=0$ has three distinct real roots, then the range of the real number $a$ is ( ).
(A) $(-16,16)$
(B) $[-16,16]$
(C) $(-\infty,-8)$
(D) $(8,+\infty)$
(5) A Hint: Let $f(x)=x^{3}-12 x+a$, then $f^{\prime}(x)=3 x^{2}-12$,
$$
f^{\prime}(x)=3 x^{2}-12=0 \Rightarrow x= \pm 2 .
$$

To make $f(x)=0$ have three different real roots, it is necessary that $f(2) f(-2)<0$, i.e., $(a-16)(a+16)<0$, which means $-16<a<16$.