What is $\gcd(a^2-1, a^3-1)$?

Question

What is $\gcd(a^2-1, a^3-1)$? Is it $1$? The exponents seem to follow the pattern of $\gcd(a, a+1)$.

Answer 1

$a^3-1 = (a-1)(a^2+a+1) = (a-1)(a(a+1)+1)$

$a^2-1=(a-1)(a+1)$

${\rm GCD}(a+1,a(a+1)+1)={\rm GCD}(a+1,1)=1$ so that ${\rm GCD} (a^2-1,a^3-1)=a-1$