6. In $\triangle A B C$, $A B=A C=m, B C=n$, $\angle B A C=160^{\circ}$. Then the relationship between $n^{3}-\sqrt{3} m^{3}$ and $3 m^{2} n$ is $n^{3}-\sqrt{3} m^{3}(\quad) 3 m^{2} n$.
(A) greater than
(B) less than
(C) equal to
(D) uncertain
6. C.

From $A B=A C$, we get
$$
\angle B=\angle A C B=\frac{180^{\circ}-\angle B A C}{2}=10^{\circ} \text {. }
$$

Construct $\angle B A D=\angle B$ inside $\angle B A C$, with point $D$ on side $B C$, and draw $C E \perp D A$ at point $E$. Then
$$
\begin{array}{l}
A D=B D, \\
\angle D A C=\angle B A C-\angle B A D=150^{\circ} .
\end{array}
$$
$$
\text { Hence } \angle C A E=180^{\circ}-\angle D A C=30^{\circ} \text {. }
$$

In Rt $\triangle A C E$,
$$
\begin{array}{l}
C E=\frac{1}{2} A C=\frac{1}{2} m, \\
A E=A C \cos \angle C A E=\frac{\sqrt{3}}{2} m .
\end{array}
$$

Therefore, $\triangle A B D \backsim \triangle C B A$
$$
\Rightarrow \frac{B D}{B A}=\frac{A B}{C B} \Rightarrow B D=\frac{A B^{2}}{C B}=\frac{m^{2}}{n} \text {. }
$$

In Rt $\triangle C D E$,
$$
\begin{array}{l}
D C^{2}=D E^{2}+C E^{2} \\
\Rightarrow(B C-B D)^{2}=(A D+A E)^{2}+C E^{2} \\
\Rightarrow\left(n-\frac{m^{2}}{n}\right)^{2}=\left(\frac{m^{2}}{n}+\frac{\sqrt{3}}{2} m\right)^{2}+\frac{1}{4} m^{2} \\
\Rightarrow n^{3}-\sqrt{3} m^{3}=3 m^{2} n .
\end{array}
$$