Triangle with 60 degree show

Question

I do not understand how to do the following problem. I have attempted to use cos rule, sin rule and even constructions to make use of the 60 degree information but i still cant reach the answer.

Answer 1

By the Law of Cosines $$c^2=a^2+b^2-ab$$ hence \begin{align*} &\frac{3}{a+b+c}-\left(\frac{1}{a+c}+\frac{1}{b+c}\right)\\[4pt] =\;&\frac{3(a+c)(b+c)-(a+b+c)(b+c)-(a+b+c)(a+c)}{(a+b+c)(a+c)(b+c)}\\[4pt] =\;&\frac{(3ab+3ac+3bc+3c^2)-(ab+ac+2bc+b^2+c^2)-(ab+bc+2ac+a^2+c^2)}{(a+b+c)(a+c)(b+c)} \\[4pt] =\;&\frac{c^2-(a^2+b^2-ab)}{(a+b+c)(a+c)(b+c)}\\[4pt] =\;&\;0 \\[4pt] \end{align*} As to why the simplification above was destined to work, the idea is that the Law of Cosines, applied to the angle $C$, yields an if-and-only-if condition for $C=60^{\circ}$, hence, if $C=60^{\circ}$ implies some other identity relating $a,b,c$, we would expect it to be an algebraic consequence of the Law of Cosines.