Ta có: a^3+ b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) = 0
=>  a^3 + b^3 + c^3 = 3abc
=> A = a^2/bc + b^2/ac + c^2/ab
         = a^3/abc + b^3/abc + c^3/abc
         = (a^3 + b^3 + c^3)/abc
         = 3abc / abc
         = 3
Vậy A = 3