A = \(n^3-4n^2+6n-4=\left(n-2\right)\left(n^2-2n+2\right)\)
A là số nguyên tố
\(\Leftrightarrow\left[{}\begin{matrix}n-2=1\\n^2-2n+2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}n=3\left(nhan\right)\\n=1\left(loai\right)\end{matrix}\right.\)
Vậy n = 3