Từ giả thiết suy ra : ab + bc + ca = 0
Ta có : (b + c)/a = (ab + ac)/a^2 = (-bc)/a^2 = (-abc)/a^3
Tương tự : (a + c)/b = (-abc)/b^3 và (a + b)/c = (-abc)/c^3
Suy ra : M = -abc.(1/a^3 + 1/b^3 + 1/c^3)
Mặt khác : 1/a + 1/b + 1/c = 0
=> (1/a + 1/b)^3 = (-1/c)^3
=> 1/a^3 + 1/b^3 + 3.1/ab.(1/a + 1/b) = -1/c^3
=> 1/a^3 + 1/b^3 + 1/c^3 + 3.1/ab.(-1/c) = 0
=> 1/a^3 + 1/b^3 + 1/c^3 = 3/(abc)
Suy ra : M = -abc.3/(abc) = -3
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=0\text{ và: }ab+bc+ac=0\)
=>\(1+\frac{b+c}{a}+1+\frac{a+c}{b}+1+\frac{b+a}{c}=0\text{ và: }\hept{\begin{cases}ab+bc=-ca\\bc+ca=-ab\\ac+ab=-bc\end{cases}}\)
=>\(\frac{ab+ac}{a^2}+\frac{ab+bc}{b^2}+\frac{bc+ac}{c^2}=-3\)
=>\(-\frac{bc}{a^2}+\frac{-ac}{b^2}+\frac{-ab}{c^2}=-3\)
=>\(\frac{bc}{a^2}+\frac{ac}{b^2}+\frac{ab}{c^2}=3\)
