\(gt\Rightarrow\frac{ab+bc+ca}{abc}=0\) \(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
\(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
\(\Rightarrow x^3+y^3+z^3=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-czx\right)+3xyz\)
+ \(A=\frac{abc}{a^3}+\frac{abc}{b^3}+\frac{abc}{c^3}\)
\(=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\)
\(=abc\left[\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}-\frac{1}{ab}-\frac{1}{bc}-\frac{1}{ca}\right)+\frac{3}{abc}\right]\)
\(=abc\cdot\frac{3}{abc}=3\)
Ta có:
ab + ac + bc = 0
\(\Rightarrow\) \(\frac{ab+ac+bc}{abc}=0\)
\(\Rightarrow\) \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
Đặt \(\frac{1}{a}=x;\) \(\frac{1}{b}=y;\) \(\frac{1}{c}=z\)
Mà x + y + z = 0
=> x3 + y3 + z3 = 3xyz (Tự chứng minh nhé bạn, nếu không chứng minh được thì bình luận nhé!)
\(\Rightarrow\) \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)
Ta có:
\(A=\frac{bc}{a^2}+\frac{ac}{b^2}+\frac{ab}{c^2}\)
\(A=\frac{abc}{a^3}+\frac{abc}{b^3}+\frac{abc}{c^3}\)
\(A=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\)
\(A=abc.\frac{3}{abc}\)
\(A=3\)
