determine the gcd of $n^3+3n^2-5 $ and $ n+2$

Question

Hello I have to determine the gcd of $n^3+3n^2-5 $ and $ n+2$, but I can’t do it because I know we have to eliminate the $n$. I tried to eliminate $n^3 $ but I’m stuck for the rest.

Answer 1

$$n^3 + 3n^2 - 5 - n^2 (n+2) = n^2 - 5$$ and $$n^2 - 5 - n(n+2) = -2n-5$$ Why did I do this? $\gcd(f(x),g(x)) = \gcd(f(x) - \lambda g(x), g(x))$ where $\lambda$ is an integer. So, keeping the above two equations in mind we have $$\gcd(n^3 + 3n^2 - 5, n+2) = \gcd(-2n-5, n+2) = 1$$ i.e. the two are co-prime (the greatest common divisor is $1$).

Answer 2

Hint: $n^3+3n^2-5=(n+2)^3-3(n+2)^2-1$.

Answer 3

Let $g$ be their gcd. Then $$n^3 +3n^2 -5 = ga \\ n+2 = gb $$ Subtract the first from the cube of the second: $$3n^2 +12n +13 = g^3b^3 -ga $$ Subtract from this thrice the square of the second equation: $$1 = g^3b^3 -ga-3g^2b^2 = g(g^2 b^3 -a-3gb^2)$$ This implies $$g=1$$