Từ đề bài \(\Rightarrow a;b;c\ne0\)
Đặt \(A=\dfrac{b-c}{a}+\dfrac{c-a}{b}+\dfrac{a-b}{c}\)
\(\Rightarrow Q=A.\dfrac{a}{b-c}+A.\dfrac{b}{c-a}+A\dfrac{c}{a-b}\)
Ta có \(\dfrac{a}{b-c}.A=\dfrac{a}{b-c}\left(\dfrac{b-c}{a}+\dfrac{c^2-b^2+ab-ac}{bc}\right)\)
\(=\dfrac{a}{b-c}\left(\dfrac{b-c}{a}+\dfrac{\left(b-c\right)\left(a-b-c\right)}{bc}\right)=1+\dfrac{2a^3}{abc}\) (do \(a-b-c=2a\))
Tương tự: \(A.\dfrac{b}{c-a}=1+\dfrac{2b^3}{abc};A.\dfrac{c}{a-b}=1+\dfrac{2c^3}{abc}\)
\(\Rightarrow Q=3+\dfrac{2}{abc}\left(a^3+b^3+c^3\right)\)
Mà do \(-c=a+b\):
\(a^3+b^3+c^3=\left(a+b\right)\left(\left(a+b\right)^2-3ab\right)+c\left(c^2-3ab\right)+3abc\)
\(=-c\left(c^2-3ab\right)+c\left(c^2-3ab\right)+3abc=3abc\)
\(\Rightarrow Q=3+\dfrac{2}{abc}.3abc=3+6=9\)
