Đáp án:
$\begin{array}{l}
a)2{n^4} + {n^3} - 3{n^2} - 3n + 6\\
= 2{n^4} - 6{n^2} + {n^3} - 2n\\
+ 3{n^2} - 9 - n + 15\\
= 2{n^2}\left( {{n^2} - 3} \right) + n\left( {{n^2} - 3} \right)\\
+ 3\left( {{n^2} - 3} \right) - n + 15\\
= \left( {{n^2} - 3} \right)\left( {2{n^2} + n + 3} \right) - n + 15\\
Do:\left( {{n^2} - 3} \right)\left( {2{n^2} + n + 3} \right) \vdots \left( {{n^2} - 3} \right)\\
\Rightarrow - n + 15 = 0\\
\Rightarrow n = 15\\
Vậy\,n = 15\\
b){n^3} - n + 5\\
= {n^3} - {n^2} + {n^2} - n + 5\\
= {n^2}\left( {n - 1} \right) + n\left( {n - 1} \right) + 5\\
= \left( {n - 1} \right)\left( {{n^2} + n} \right) + 5\\
DO:\left( {n - 1} \right)\left( {{n^2} + n} \right) \vdots n - 1\\
\Rightarrow 5 \vdots n - 1\\
\Rightarrow \left( {n - 1} \right) \in \left\{ { - 5; - 1;1;5} \right\}\\
\Rightarrow n \in \left\{ { - 4;0;2;6} \right\}
\end{array}$
