Ta có:
\(a^2-b=b^2-c=c^2-a\Rightarrow\hept{\begin{cases}a^2-b^2=b-c\\b^2-c^2=c-a\\c^2-a^2=a-b\end{cases}}\)\(\Rightarrow\hept{\begin{cases}a+b=\frac{b-c}{a-b}\\b+c=\frac{c-a}{b-c}\\c+a=\frac{a-b}{c-a}\end{cases}}\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=\frac{b-c}{a-b}.\frac{c-a}{b-c}.\frac{a-b}{c-a}=1\)