\(\Leftrightarrow\left\{{}\begin{matrix}x-a=b-y\\x^2-a^2=b^2-y^2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-a=b-y\\\left(x-a\right)\left(x+a\right)=\left(b-y\right)\left(b+y\right)\left(1\right)\end{matrix}\right.\)
- Nếu \(x-a=b-y=0\Leftrightarrow\left[{}\begin{matrix}x=a\\y=b\end{matrix}\right.\)
\(\Rightarrow M=a^8+b^8\)
- Nếu \(x-a=b-y\ne0\)
Từ (1) \(\Rightarrow x+a=b+y\)
\(\Rightarrow\left\{{}\begin{matrix}x-a=b-y\\x+a=b+y\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2x=2b\\2y=2a\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=b\\y=a\end{matrix}\right.\)
\(\Rightarrow M=x^8+y^8=a^8+b^8\)
Vậy trong mọi trường hợp thì \(M=a^8+b^8\)
