6. Given the three sides of $\triangle A B C$ are $a, b, c$. If $a+b+c=16$, then
$$
b^{2} \cos ^{2} \frac{C}{2}+c^{2} \cos ^{2} \frac{B}{2}+2 b c \cos \frac{B}{2} \cdot \cos \frac{C}{2} \cdot \sin \frac{A}{2}
$$
$=$
6. 64 .

Let the circumradius of $\triangle ABC$ be $R$. Then
$$
\begin{array}{l} 
b^{2} \cos ^{2} \frac{C}{2}+c^{2} \cos ^{2} \frac{B}{2}+2 b c \cos \frac{B}{2} \cdot \cos \frac{C}{2} \cdot \sin \frac{A}{2} \\
= 16 R^{2} \cos ^{2} \frac{B}{2} \cdot \cos ^{2} \frac{C}{2}\left(\sin ^{2} \frac{B}{2}+\sin ^{2} \frac{C}{2}+\right. \\
\left.2 \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2}\right) \\
= 16 R^{2} \cos ^{2} \frac{B}{2} \cdot \cos ^{2} \frac{C}{2}\left(\frac{1-\cos B}{2}+\right. \\
\left.\frac{1-\cos C}{2}+\frac{\cos A+\cos B+\cos C-1}{2}\right) \\
= 16 R^{2} \cos ^{2} \frac{A}{2} \cdot \cos ^{2} \frac{B}{2} \cdot \cos ^{2} \frac{C}{2} \\
= R^{2}(\sin A+\sin B+\sin C)^{2} \\
= \frac{1}{4}(a+b+c)^{2}=64 .
\end{array}
$$