\(=n^3+n^3+3n^2+3n+1+n^3+6n^2+12n+8\)
\(=3n^3+9n^2+15n+9\)
\(=3\left(n^3+3n^2+5n+3\right)\)
\(=3\left(n^3+n^2+2n^2+2n+3n+3\right)\)
\(=3\left(n+1\right)\left(n^2+2n+3\right)\)
\(=3n\left(n+1\right)\left(n+2\right)+9\left(n+1\right)\)
Vì n;n+1;n+2 là ba số liên tiếp
nên \(3n\left(n+1\right)\left(n+2\right)⋮3\cdot3=9\)
=>A chia hết cho 9