In a triangle $ABC$, if $A=60°$ then the value of $\left(1+\frac{\mathrm{a}}{\mathrm{c}}+\frac{\mathrm{b}}{\mathrm{c}}\right)\left(1+\frac{\mathrm{c}}{\mathrm{b}}-\frac{\mathrm{a}}{\mathrm{b}}\right)$ is 

Given that in triangle $ABC$

The measure of angle $A$ is $60°$

We know that from the half angle formule

$\sin \frac{A}{2}=\sqrt{\frac{s\left(s-a\right)}{bc}}$

Squaring on both  sides

$\begin{array}{rcl}Si{n}^2\left(\frac{A}{2}\right)& =& \frac{2s\left(2s-2a\right)}{4bc}\\ & =& \frac{\left(a+b+c\right)\left(b+c-a\right)}{4bc}\\ 4\left(\frac{3}{4}\right)& =& \left(1+\frac{a}{c}+\frac{b}{c}\right)\left(1+\frac{c}{b}-\frac{a}{b}\right)\end{array}$

Therefore, $\left(1+\frac{a}{c}+\frac{b}{c}\right)\left(1+\frac{c}{b}-\frac{a}{b}\right)=\boxed{\mathbf{3}}$