Ta có: \(ab+bc+ac=0\) và \(abc\ne0\)
nên \(\frac{ab+bc+ac}{abc}=0\), tức là \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
Nếu \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\) thì \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=3.\frac{1}{a}.\frac{1}{b}.\frac{1}{c}=\frac{3}{abc}\)
(bạn tham khảo cách chứng minh tại link sau: http://olm.vn/hoi-dap/question/373691.html)
Do đó: \(A=\frac{bc}{a^2}+\frac{ac}{b^2}+\frac{ab}{c^3}=\frac{abc}{a^3}+\frac{abc}{b^3}+\frac{abc}{c^3}=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=abc.\frac{3}{abc}=3\)
với \(a,b,c\ne0\)
\(ab+bc+ca=0\)
\(\Leftrightarrow ab+bc=-ac\)
\(\Leftrightarrow a^3b^3+b^3c^3+3ab^2c\left(ab+bc\right)=-a^3c^{3 }\)
\(\Leftrightarrow a^3b^3+b^3c^3+c^3a^3=3a^2b^2c^2\)
\(\Leftrightarrow\frac{a^3b^3+b^3c^3+c^3a^3}{a^2b^2c^2}=\frac{3a^2b^2c^2}{a^2b^2c^2}\)
\(\Leftrightarrow A=3\)