Is it true that $\forall a,b\in \mathbb{Z}$, $\gcd(a^3, b^3)=\gcd(a,b)^3$? I cannot find a counterexample, nor have I been able to finish a proof. One thing I tried was:
$\gcd(a^3, b^3)= \gcd(a^3, b^3-a^3)=\gcd(a^3, (b-a)(b^2+ba+a^2))$ but I don't see how this helps.
$a=gcd(a,b)j$
$b=gcd(a,b)k$
with $gcd(j,k)=1$
$a^3=gcd(a,b)^3j^3$
$b^3=gcd(a,b)^3k^3$
$gcd(j^3, k^3)=1$
Can you see it now?