If  $a=5,b=4\text{\hspace{0.17em}and\hspace{0.17em}}\cos (A-B)=\frac{31}{32}$, then  $\text{'}c\text{'}=$

$c^2=a^2+b^2-2ab\cos C$

$=25+16-\mathrm{2.5.4}\cos C$

$c^2=41-40\cos C\to \left(1\right)$

$\tan \left(\frac{A-B}{2}\right)=\frac{a-b}{a+b}\cot \frac{C}{2}$

$\sqrt{\frac{1-\cos (A-B)}{1+\cos (A-B)}}=\frac{a-b}{a+b}\cot \frac{C}{2}$

$\sqrt{\frac{1-\frac{31}{32}}{1+\frac{31}{32}}}=\frac{5-4}{5+4}\cot \frac{C}{2}$

$\Rightarrow \cot \frac{C}{2}=\frac{3}{\sqrt{7}}$

$\Rightarrow \tan \frac{C}{2}=\frac{\sqrt{7}}{3}$

$c^2=41-40\frac{1-{\tan }^2\frac{C}{2}}{1+{\tan }^2\frac{C}{2}}$

$c^2=41-40\left(\frac{1-\frac{7}{9}}{1+\frac{7}{9}}\right)$

$=41-40\left(\frac{2}{16}\right)$

$c^2=41-5=36\Rightarrow c=6$