Question

cho \(a^2-b=b^2-c=c^2-a\)

tính giá trị của biểu thức \(P=\left(a+b\right)\left(b+c\right)\left(c+a\right)\)

Answer 1

\(\Rightarrow\hept{\begin{cases}a^2-b-\left(b^2-c\right)=0\\b^2-c-\left(c^2-a\right)=0\\c^2-a-\left(a^2-b\right)=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}\left(a-b\right)\left(a+b\right)=b-c\\\left(b-c\right)\left(b+c\right)=c-a\\\left(c-a\right)\left(c+a\right)=a-b\end{cases}}\)

\(\Rightarrow\left(a-b\right)\left(a+b\right)\left(b-c\right)\left(b+c\right)\left(c-a\right)\left(c+a\right)=\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

\(\Leftrightarrow\orbr{\begin{cases}\left(a-b\right)\left(b-c\right)\left(c-a\right)=0\\\left(a+b\right)\left(b+c\right)\left(c+a\right)=1\end{cases}}\)