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Solution:

We have,

$f(x)=x^{3}+a x^{2}+b x+c$

$\Rightarrow f^{\prime}(x)=3 x^{2}+2 a x+b$

As, $f(x)$ is maximum at $x=-1$ and minimum at $x=3$.

So, $f(-1)=0$ and $f(3)=0$

$\Rightarrow 3(-1)^{2}+2 a(-1)+b=0$ and $3(3)^{2}+2 a(3)+b=0$

$\Rightarrow 3-2 a+b=0 \quad \ldots$ (i)

and $27+6 a+b=0 \quad \ldots$ (ii)

(ii) - (i), we get

$27-3+6 a+2 a=0$

$\Rightarrow 8 a=-24$

$\Rightarrow a=-3$

Substituting $a=-3$ in (i), we get

$3-2(-3)+b=0$

$\Rightarrow 3+6+b=0$

$\Rightarrow b=-9$

And, $c \in \mathbf{R}$