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 $\gcd (bc, a)=k$.
 A: Hint $\rm\ \ (a,bc)\ =\ (a\:(1,c),bc)\ =\ (a,ac,bc)\ =\ (a,(a,b)\:c)\ \ [\:=\ (a,c)\ \ if\ \  (a,b)=1\ ] $ 
The above proof uses only basic gcd laws (associative, commutative, distributive) - see here.
Alternatively, if your prefer to use Bezout's identity, consider the following
$$\rm \begin{array}{rl}
\rm (a,b)\:(a,c) &=&\rm (a\:(a,b,c),\:bc)\\
 \Rightarrow\quad\rm (a,c) &=&\rm (a,\:bc)\quad if\quad (a,b)= 1
\end{array}\ \ $$
The first identity above can easily be recast into Bezout form,namely
$$\begin{eqnarray}
 \rm(b,\  a)\ \ \ &\rm\!\!\! (c,\ \ a)&=&\rm\ (bc,\ &\rm a\:(a\:,\ &\rm b\:,\ &\rm c))\qquad \\
\rm (br\!\!+\!\!as)\:&\rm(ct\!\!+\!\!au) &=&\rm\ \ bc\:(rt)\!\!+\!\!&\rm a\:(asu\!\!+\!\!&\rm bru\!\!+\!\!&\rm cst)
\end{eqnarray}$$
For further discussion see my post here.
A: 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 $gcd(a,bc)=k.$
