\(A=\dfrac{\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc}{2\left(a^2+b^2+c^2-ab-ac-bc\right)}\)
\(=\dfrac{\left(a+b+c\right)\cdot\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)}{2\left(a^2+b^2+c^2-ab-ac-bc\right)}\)
\(=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)}{2\left(a^2+b^2+c^2-ab-ac-bc\right)}\)
=3/2