14. Given that the function $f(x)$ satisfies the following five conditions:
(1) The domain of $f(x+1)$ is $[-5,3]$;
(2) $f(x)+f(-x)=0$;
(3) $f(-1)=0$;
(4) It is monotonically decreasing on $[-4,0)$: (5) It has no maximum value.
Then the solution set of the inequality $x^{3} f(x) \leqslant 0$ is $\qquad$.
14. From (1), we know that the domain of $f(x)$ is $[-4,4]$, from (2) we know that $f(x)$ is an odd function, from (1) and (2) we know that $f(0)=0$, from (3) we know that the graph of the function passes through the point $(-1,0)$, from (2) and (3) we know that the graph of the function passes through the point $(1,0)$.
From the above, we can conclude that when $x \in[-4,-1) \cup(0,1)$, $f(x)>0$;
when $x=-1,0,1$, $f(x)=0$;
when $x \in(-1,0) \cup(1,4]$, $f(x)<0$,
so the solution set of $x^{3} f(x) \leqslant 0$ is $\{x \mid-4 \leqslant x \leqslant-1$ or $x=0$ or $1 \leqslant x \leqslant 4\}$.