\(\frac{1}{ab+a+1}+\frac{b}{bc+b+1}+\frac{1}{abc+bc+b}\)
\(=\frac{1}{ab+a+1}+\frac{b}{bc+b+1}+\frac{1}{1+bc+b}\)
\(=\frac{1}{ab+a+1}+\frac{a.b}{a.\left(bc+b+1\right)}+\frac{1.a}{a.\left(1+bc+b\right)}\)
\(=\frac{1}{ab+a+1}+\frac{ab}{abc+ab+a}+\frac{a}{a+abc+ab}\)
\(=\frac{1}{ab+a+1}+\frac{ab}{ab+a+1}+\frac{a}{ab+a+1}=\frac{ab+a+1}{ab+a+1}=1\)