6:
a: a+b+c=0
=>a=-b-c; b=-a-c; c=-a-b
=>a^2=(b+c)^2; b^2=(a+c)^2; c^2=(a+b)^2
=>4bc-a^2=4bc-(b+c)^2=-(b-c)^2; 4ac-b^2=-(a-c)^2; 4ab-c^2=-(a-b)^2
\(M=\dfrac{-\left(b-c\right)^2}{bc+2a^2}\cdot\dfrac{-\left(a-c\right)^2}{ca+2b^2}\cdot\dfrac{-\left(a-b\right)^2}{ab+2c^2}\)
\(=-\dfrac{\left(b-c\right)^2}{2b^2+4bc+2c^2+bc}\cdot\dfrac{\left(a-c\right)^2}{2a^2+4ac+2c^2+ac}\cdot\dfrac{\left(a-b\right)^2}{2a^2+4ab+2b^2+ab}\)
\(=\dfrac{-\left(b-c\right)^2}{2b\left(b+2c\right)+c\left(b+2c\right)}\cdot\dfrac{\left(a-c\right)^2}{2a\left(2c+a\right)+c\left(2c+a\right)}\cdot\dfrac{\left(a-b\right)^2}{2a\left(2b+a\right)+b\left(2a+b\right)}\)
\(=\dfrac{-\left(b-c\right)^2}{\left(b+2c\right)\left(2b+c\right)}\cdot\dfrac{\left(a-c\right)^2}{\left(2a+c\right)\left(2c+a\right)}\cdot\dfrac{\left(a-b\right)^2}{\left(2b+a\right)\left(2a+b\right)}\)
\(=\dfrac{-\left(b-c\right)^2}{\left(-a-c+2c\right)\left(2b-a-b\right)}\cdot\dfrac{\left(a-c\right)^2}{\left(2a-a-b\right)\left(2c-b-c\right)}\cdot\dfrac{\left(a-b\right)^2}{\left(2b-b-c\right)\left(2a-a-c\right)}\)
\(=\dfrac{-\left(b-c\right)^2}{\left(c-a\right)\left(b-a\right)}\cdot\dfrac{\left(a-c\right)^2}{\left(a-b\right)\left(c-b\right)}\cdot\dfrac{\left(a-b\right)^2}{\left(b-c\right)\left(a-c\right)}\)
\(=\dfrac{-\left(b-c\right)^2\left(a-c\right)^2\left(a-b\right)^2}{-\left(a-b\right)^2\cdot\left[-\left(a-c\right)^2\right]\cdot\left[-\left(b-c\right)^2\right]}\)
=1
b: \(N=\dfrac{a+b}{b}\cdot\dfrac{b+c}{c}\cdot\dfrac{a+c}{a}=\dfrac{\left(-c\right)\cdot\left(-a\right)\cdot\left(-b\right)}{abc}=-1\)
