\(P=\frac{1}{a^2+\left(b+c\right)\left(b-c\right)}+\frac{1}{b^2+\left(c+a\right)\left(c-a\right)}+\frac{1}{c^2+\left(a+b\right)\left(a-b\right)}\)
\(=\frac{1}{a^2-a\left(b-c\right)}+\frac{1}{b^2-b\left(c-a\right)}+\frac{1}{c^2-c\left(a-b\right)}\)
\(=\frac{1}{a\left(a+c-b\right)}+\frac{1}{b\left(b+a-c\right)}+\frac{1}{c\left(c+b-a\right)}\)
\(=\frac{1}{a\left(-2b\right)}+\frac{1}{b\left(-2c\right)}+\frac{1}{c\left(-2a\right)}=\frac{-\left(a+b+c\right)}{2abc}=0\)
Ta có : a + b + c = 0
=> a + b = - c => a2 + 2ab + b2 = c2
=> a2 + b2 - c2 = - 2ab
Tương tự : b2 + c2 - a2 = - 2bc
c2 + a2 - b2 = - 2ac
=> \(P=\frac{-1}{2ab}-\frac{1}{2bc}-\frac{1}{2ac}=\frac{-c-a-b}{2abc}=\frac{-\left(a+b+c\right)}{2abc}=0\)
