Ta có:
\(n^3+6n^2+11n+6=(n+1)(n+3)\)
Để
\(n^3+6n^2+11n+6\) chia hết cho \(3\) thì
\((n+1)\) chia hết cho \(3\) (1).
hoặc \((n+3)\) chia hết cho \(3\) (2).
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`(1)`
\(\Leftrightarrow (n+1)=B(3)\)
\(\Leftrightarrow n+1=3k, k\in\mathbb N\)
\(\Leftrightarrow n=3k-1,k\in\mathbb N\).
`(2)`
\(\Leftrightarrow (n+3)=B(3)\)
\(\Leftrightarrow n+3=3k, k\in\mathbb N\)
\(\Leftrightarrow n=3k-3, k\in\mathbb N\)
\(\Leftrightarrow n=3k, k\in\mathbb N\text{*}\).
