Ta có: \(a+b+c=0\Leftrightarrow a+b=-c\)
\(\Leftrightarrow\left(a+b\right)^3+c^3=0\)
\(\Leftrightarrow a^3+b^3+c^3+3ab\left(a+b\right)=0\)
\(\Leftrightarrow a^3+b^3+c^3-3abc=0\Leftrightarrow a^3+b^3+c^3=3abc\left(\circledast\right)\)
\(A=\dfrac{a^2}{bc}+\dfrac{b^2}{ac}+\dfrac{c^2}{ab}=\dfrac{1}{abc}\left(a^3+b^3+c^3\right)=\dfrac{3abc}{abc}=3\left(từ\circledast\right)\)