A= n³ + 5n

= n[n² + 5]

1. Prove that A divisible by 2

- If n is a even => A divisible by 2

- If n is a odd => n² is a odd => n² + 5 divisible 2 => A divisible by 2

So A divisible by 2

2. Prove that: A divisible by 3

- If n divisible by 3 => A divisible by 3

- If n not divisible by 3 => n² \(\equiv1\left(mod3\right)\)

=> n² + 5 divisible by 3

=> A divisible by 3

So A divisible by 3

But \(\left(2,3\right)=1\)

=> A divisible by 2.3 = 5

Or A divisible by 6

\(A=n^3+5n=n^3-n+6n=n\left(n^2-1\right)+6n=\left(n-1\right)n\left(n+1\right)+6n\)

(n-1).n.(n+1) is the product of the 3 consecutive numbers ;so : (n-1).n.(n+1) divisible 6 

=> A divisible 6 (đpcm)