3. When $x=\frac{1}{\sqrt{3}-2}$, the value of the function $f(x)=x^{3}+4 x^{2}-$ $2 x-6$ is ( ).
(A) $-3 \sqrt{3}$
(B) $3 \sqrt{3}$
(C) $-6 \sqrt{3}$
(D) $6 \sqrt{3}$
3. B.

Notice that $x=\frac{1}{\sqrt{3}-2}=-(2+\sqrt{3})$, i.e., $x+2=-\sqrt{3}$.
Therefore, when $x=\frac{1}{\sqrt{3}-2}$,
$$
\begin{array}{l}
f(x)=x(x+2)^{2}-6(x+1) \\
\quad=3 x-6(x+1)=-3(x+2)=3 \sqrt{3} .
\end{array}
$$