Ta có:
\(M=\dfrac{bc}{a^2}+\dfrac{ac}{b^2}+\dfrac{ab}{c^2}=\dfrac{abc}{a^3}+\dfrac{abc}{b^3}+\dfrac{abc}{c^3}\)
\(M=abc\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)\)
Áp dụng hằng đẳng thức mở rộng ta có:
\(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}-\dfrac{1}{ab}-\dfrac{1}{bc}-\dfrac{1}{ac}\right)+\dfrac{3}{abc}\)
Hay: \(M=abc\left[\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}-\dfrac{1}{ab}-\dfrac{1}{bc}-\dfrac{1}{ac}\right)+\dfrac{3}{abc}\right]\)
\(M=abc\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}-\dfrac{1}{ab}-\dfrac{1}{bc}-\dfrac{1}{ac}\right)+\dfrac{3abc}{abc}\)
\(M=0+3=3\)
