Ta có :
\(ab-ac+bc=c^2-1\)
\(\Leftrightarrow a\left(b-c\right)+c\left(b-c\right)=-1\)
\(\Leftrightarrow\left(b-c\right)\left(a+c\right)=-1\)
\(\Leftrightarrow b-c=1;a+c=-1\) hoặc \(b-c=-1;a+c=1\)
\(\Leftrightarrow\left(b-c\right)+\left(a+c\right)=1+\left(-1\right)\)
\(\Leftrightarrow b+a=0\)
\(\Leftrightarrow a;b\) là hai số đối nhau
\(\Leftrightarrow a:b=-1\)