Với \(a+b+c=0;abc\ne0\), ta có:
\(A=\frac{\left(a^2+b^2-c^2\right)\left(b^2+c^2-a^2\right)\left(c^2+a^2-b^2\right)}{10a^2b^2c^2}\) (*)
Ta có: \(a+b+c=0\)
\(\Leftrightarrow a+b=-c\)
\(\Rightarrow\left(a+b\right)^2=\left(-c\right)^2\)
\(a^2+2ab+b^2=c^2\)
Suy ra \(a^2+b^2-c^2=a^2+b^2-\left(a^2+2ab+b^2\right)\)
\(=a^2+b^2-a^2-2ab-b^2\)
\(=-2ab\)
Vậy \(a^2+b^2-c^2=-2ab\) (1)
Chứng minh tương tự: \(b^2+c^2-a^2=-2bc\) (2)
và \(c^2+a^2-b^2=-2ac\) (3)
Thay (1), (2), (3) vào (*), ta được:
\(A=\frac{\left(-2ab\right)\left(-2bc\right)\left(-2ac\right)}{10a^2b^2c^2}=\frac{-8a^2b^2c^2}{10a^2b^2c^2}=\frac{-4}{5}\)
