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Gọi $A=\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}$

$\rightarrow A.\dfrac{c}{a-b}=1+\dfrac{c}{a-b}.(\dfrac{b-c}{a}+
\dfrac{c-a}{b})$

$=1+\dfrac{c}{a-b}.\dfrac{b^2-bc+ac-a^2}{ab}=1+\dfrac{c}{a-b}.
\dfrac{(a-b)(c-a-b)}{ab}=1+\dfrac{c(2c-a-b-c)}{ab}=1+\dfrac{2c^2}
{ab}=1+\dfrac{2c^3}{abc}$

Tương tự $A.\dfrac{a}{b-c}=1+\dfrac{2a^3}{abc} \\ A.\dfrac{b}
{c-a}=1+\dfrac{2b^3}{abc}$

$\rightarrow M=3+\dfrac{2(a^3+b^3+c^3)}{abc}$

Ta có $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0 \rightarrow
a^3+b^3+c^3=3abc$

$\rightarrow M=3+\dfrac{6abc}{abc}=9$

Vậy $M=9$


 
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