How to prove: if $a$ is an even integer, $\gcd(a^3 - 1, a + 1) = 1$

Question

I have very little idea of how to tackle this question. I know if $a$ is even, $a = 2L$, for some $L$ in the integer set.

Answer 1

Actually, since $a$ is even, $\gcd(a + 1, a - 1) = 1$.

Since $a^3-1 =(a-1)(a^2+a+1) $, $\gcd(a^3-1, a+1) =\gcd(a^2+a+1, a+1) =\gcd(a^2, a+1) =1 $.