(15) The cubic function $f(x)=4 x^{3}+a x^{2}+b x+c$ (where $a, b, c \in \mathbf{R}$) satisfies: when $-1 \leqslant x \leqslant 1$, $-1 \leqslant f(x) \leqslant 1$. Find all possible values of $a$, $b$, and $c$.

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15 Given the problem, when $x= \pm 1, \pm \frac{1}{2}$, we have $-1 \leqslant f(x) \leqslant 1$. Therefore,
$$
\begin{array}{c}
-1 \leqslant 4+a+b+c \leqslant 1, -1 \leqslant 4-a+b-c \leqslant 1, \\
-1 \leqslant \frac{1}{2}+\frac{a}{4}+\frac{b}{2}+c \leqslant 1, -1 \leqslant \frac{1}{2}-\frac{a}{4}+\frac{b}{2}-c \leqslant 1 .
\end{array}
$$

Adding the first two inequalities, we get $-2 \leqslant 8+2 b \leqslant 2$, thus $b \leqslant-3$. Adding the last two inequalities, we get $-2 \leqslant 1+b \leqslant 2$, thus $b \geqslant-3$. Therefore, $b=-3$. Substituting $b=-3$ into the aforementioned inequalities, we get $a+c=0, \frac{a}{4}+c=0$, thus $a=c=0$.

Next, we prove that $f(x)=4 x^{3}-3 x$ satisfies the conditions stated in the problem. In fact, if $-1 \leqslant x \leqslant 1$, then we can let $x=\cos t, t \in \mathbf{R}$, at this time
$$
f(x)=f(\cos t)=4 \cos ^{3} t-3 \cos t=\cos 3 t .
$$

Since $-1 \leqslant \cos 3 t \leqslant 1$, it follows that $-1 \leqslant f(x) \leqslant 1$.
In summary, $a=0, b=-3, c=0$.