Ta có: $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow \frac{ab+bc+ca}{abc}=0\Leftrightarrow ab+bc+ca=0$
Mặt khác: $ab+bc+ca=0 \Rightarrow 2bc=bc-ab-ac$
Suy ra: $\frac{1}{a^{2}+2bc}=\frac{1}{a^{2}+bc-ab-ac}=\frac{1}{\left ( a-b \right )\left ( a-c \right )}=\frac{c-b}{\left ( a-b \right )\left ( b-c \right )\left ( c-a \right )}$ (1)
Tương tự ta có:
$\frac{1}{b^{2}+2ca}=\frac{a-c}{\left ( a-b \right )\left ( b-c \right )\left ( c-a \right )}$ (2)
$\frac{1}{c^{2}+2ab}=\frac{b-a}{\left ( a-b \right )\left ( b-c \right )\left ( c-a \right )}$ (3)
Từ (1); (2) và (3) suy ra: $M=\frac{c-b+a-c+b-a}{\left ( a-b \right )\left ( b-c \right )\left ( c-a \right )}+2014=2014$
Vậy M = 2014