\(A=\dfrac{a^3+b^3+c^3}{abc}=\dfrac{3abc}{abc}=3\)
Ta có :\(A=\dfrac{a^2}{bc}+\dfrac{b^2}{ac}+\dfrac{c^2}{ab}=\dfrac{a^3+b^3+c^3}{abc}\) Lại có: a3+b3+c3= [a3+3ab(a+b)+b3] +c3 - 3ab(a+b) =(a+b)3+c3-3ab(a+b) = (a+b+c)[a2+2ab+b2-ac-bc+c2] -3ab(a+b) = -3ab(a+b) = -3ab(-c) = 3abc => A=\(\dfrac{3abc}{abc}=3\)