Showing $\gcd(n^3 + 1, n^2 + 2) = 1$, $3$, or $9$ Given that n is a positive integer show that $\gcd(n^3 + 1, n^2 + 2) = 1$, $3$, or $9$. 
I'm thinking that I should be using the property of gcd that says if a and b are integers then gcd(a,b) = gcd(a+cb,b). So I can do things like decide that $\gcd(n^3 + 1, n^2 + 2) = \gcd((n^3+1) - n(n^2+2),n^2+2) = \gcd(1-2n,n^2+2)$ and then using Bezout's theorem I can get $\gcd(1-2n,n^2+2)= r(1-2n) + s(n^2 +2)$ and I can expand this to $r(1-2n) + s(n^2 +2) = r - 2rn + sn^2 + 2s$ However after some time of chasing this path using various substitutions and factorings I've gotten nowhere. 
Can anybody provide a hint as to how I should be looking at this problem?

Let $d=\gcd(n^3+1,n^2+2)$ then we have:
 $d|n^3+2n\Rightarrow\ d|2n-1\Rightarrow\ d|2n^2-n\Rightarrow\ d|2n^2+4\Rightarrow\ d|n+4\Rightarrow\ d|2n+8\Rightarrow\ d|9\Rightarrow\ d=1,3,9$
