a^3+b^3+c^3=3abc
=>(a+b)^3+c^3-3ab(a+b)-3abc=0
=>(a+b+c)(a^2+b^2+c^2-ab-ac-bc)=0
=>a^2+b^2+c^2-ab-bc-ac=0
=>2a^2+2b^2+2c^2-2ab-2bc-2ac=0
=>(a^2-2ab+b^2)+(b^2-2bc+c^2)+(a^2-2ac+c^2)=0
=>a=b=c
\(A=\left(\dfrac{b}{b}-3\right)+\left(\dfrac{b}{b}-4\right)+\left(\dfrac{c}{c}-5\right)=3-4-5=-6\)