Từ \(\left(x+1\right)^6+\left(y-1\right)^4=-z^2\)
\(\Rightarrow\left(x+1\right)^6+\left(y-1\right)^4+z^2=0\)
Thấy: \(\left\{{}\begin{matrix}\left(x+1\right)^6\ge0\forall x\\\left(y-1\right)^4\ge0\forall y\\z^2\ge0\forall z\end{matrix}\right.\)
\(\Rightarrow\left(x+1\right)^6+\left(y-1\right)^4+z^2\ge0\)
Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}\left(x+1\right)^6=0\\\left(y-1\right)^4=0\\z^2=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-1\\y=1\\z=0\end{matrix}\right.\)
Khi đó \(N=2018\cdot x^{2016}\cdot y^{2017}-\left(z-1\right)^{2018}\)
\(=2018\cdot\left(-1\right)^{2016}\cdot1^{2017}-\left(0-1\right)^{2018}\)
\(=2018-\left(-1\right)^{2018}=2018-1=2017\)