# Prove that (2a³+3a²+a)|6 , for any a belonging to N(Natural Set)

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### 1 Answer

We need to show that 6 divides `2a^3+3a^2+a` for any `a in NN`

Factoring we get `2a^3+3a^2+a=a(a+1)(2a+1)` . It suffices to show that at least 1 of the factors is even and also that one of the factors is a multiple of 3.

Consider the numbers `2a,2a+1,2a+2` :

(1) If `2a` is a multiple of 3, `a` is also. If a is even, then a is a multiple of 6 and the product above is a multiple of 6. If a is odd, then a+1 is even, and the product is a multiple of 6.

(2) If `2a+1` is a multiple of 3, then either a or a+1 is even, and the product is a multiple of 6.

(3) If `2a+2` is a multiple of 3, then a+1 is a multiple of 3. If a+1 is even, then a+1 is a multiple of 6 and the product is a multiple of 6. If a+1 is odd, then a is even and the product is a multiple of 6.

**Since this covers all possibilities, the product `a(a+1)2a+1)` is a multiple of 6 for all `a in NN` **