Prove algebraically that n^3+3n^2+2n+1 is odd for all integers n

To show this we should consider when n is odd and when n is even,If n is odd then we can find an m such that n=2m+1. Substituting n=2m+1 and expanding gives 8m^3+12m^2+6m+1+12m^2+12m+3+4m+2+1=8m^3+24m^2+22m+7.We see that 2 divides all the coeffients except the last one. So we can rearrange it as 2(4m^3+12m^2+11m)+7. This is an even number added to an odd one, so it is odd.if n is even we can find an m such that n=2msubstituting n=2m and expanding gives 8m^3+12m^2+4m+1We see that 2 divides all the coeffients except the last one. So we can rearrange it as 2(4m^3+12m^2+4m)+1This is an even number added to an odd one, so it is odd.In conclusion, since n has to be either odd or even, and since both odd and even make n^3+3n^2+2n+1 odd, we get that n^3+3n^2+2n+1 is odd for any integer n.