If $a, b, c$ are integers, $\gcd(a,b) = 1$ then $\gcd (a,bc)=\gcd(a,c)$ If $a, b, c$ and $k$ be integers, $\gcd(a,b) = 1$ and $\gcd(a, c)=k$, then $\gcd (bc, a)=k$.

Since $gcd(a,b)=1$, there exist two integers $x$ and $y$ such that $$ax+by=1\tag{1}$$
Also $gcd (a,c)=k$, there exist two integers $x_{1}$ and $y_{1}$ such that $$ax_{1}+cy_{1}=k\tag{2}$$
Now multiplying $(1)$ and $(2)$ we get,
$$a^{2}xx_{1}+acxy_{1}+bayx_{1}+bcyy_{1}=k$$
$$\Rightarrow a(axx_{1}+cxy_{1}+byx_{1})+bc(yy_{1})=k$$
It follows that $gcd(a,bc)=k.$
