Ta có: \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0\)

\( \Leftrightarrow \frac{{ab + bc + ca}}{{abc}} = 0\)

⇔ ab + bc + ca = 0

Mặt khác, \(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0\)

\( \Leftrightarrow \frac{1}{a} + \frac{1}{b} = - \frac{1}{c}\)

\( \Leftrightarrow {\left( {\frac{1}{a} + \frac{1}{b}} \right)^3} = - \frac{1}{{{c^3}}}\)

\( \Leftrightarrow \frac{1}{{{a^3}}} + \frac{1}{{{b^3}}} + 3.\frac{1}{{ab}}.\left( {\frac{1}{a} + \frac{1}{b}} \right) = - \frac{1}{{{c^3}}}\)

\( \Leftrightarrow \frac{1}{{{a^3}}} + \frac{1}{{{b^3}}} + 3.\frac{1}{{ab}}.\left( { - \frac{1}{c}} \right) = \frac{{ - 1}}{{{c^3}}}\)

\( \Leftrightarrow \frac{1}{{{a^3}}} + \frac{1}{{{b^3}}} + \frac{1}{{{c^3}}} = \frac{3}{{abc}}\)  (*)

Khi đó: \(\frac{{\left( {b + c} \right)}}{a} = \frac{{ab + ac}}{{{a^2}}} = \frac{{ - bc}}{{{a^2}}} = \frac{{ - abc}}{{{a^2}}}\)

Tương tự ta có: \(\frac{{\left( {a + b} \right)}}{c} = \frac{{ - abc}}{{{c^3}}}\); \(\frac{{\left( {a + c} \right)}}{{{b^2}}} = \frac{{ - abc}}{{{b^3}}}\).

\(M = \frac{{ - abc}}{{{a^3}}} + \frac{{ - abc}}{{{b^3}}} + \frac{{ - abc}}{{{c^3}}}\)

\( = - abc\left( {\frac{1}{{{a^3}}} + \frac{1}{{{b^3}}} + \frac{1}{{{c^3}}}} \right)\)

\( = - abc.\frac{3}{{abc}} = - 3\) (theo *)

Vậy M = −3.