a question on the cube of gcd

Question

This is an old question. In that answer,how can we get $\gcd(j^3, k^3)=1?$ What formula did he use? question

Answer 1

In the solution linked to, we have $\gcd(j,k)=1$. We want to conclude from this that $\gcd(j^3,k^3)=1$.

Suppose to the contrary that $\gcd(j^3,k^3)=d\gt 1$. Then some prime $p$ divides $d$.

So $p$ divides $j^3$, and therefore $p$ divides $j$. Similarly, $p$ divides $k$. So $p$ divides both $j$ and $k$, contradicting the fact that $\gcd(j,k)=1$.