Show that if $ \gcd(a,b)=d,\gcd(a,c)=f,\gcd(b,c)=1 \ \Rightarrow\gcd(a,bc)=df$ Show that: if $ \gcd(a,b)=d,\gcd(a,c)=f,\gcd(b,c)=1 \ \Rightarrow\ \gcd(a,bc)=df$
My work:
Let $d'=\gcd(a,bc)$, we must show that: $d'|df\  \text {and} \ df|d' $
i) Showing $d'|df$:
  $$d'|a,d'|bc,\gcd(b,c)=1\Rightarrow\ d'|b\ \  or\ \ d'|c$$
  $$if \ d'|b\Rightarrow\ d'|\gcd(a,b)=d\Rightarrow\ d'|df$$
  similarly if$\ d'|c\Rightarrow\ d'|(a,c)=f\Rightarrow\ d'|df$
ii) Showing $df|d'$
Here I stopped!!
Who can help?
 A: You want $df\mid a$ and $df\mid bc$. You know $d\mid b$ and $f\mid c$ so clearly $df\mid bc$. 
Now, $\gcd(d,f)=1$. To see that let $\gcd(d,f)=x>1$. Then $x\mid b$ and $x\mid c$, so $\gcd(b,c)>1$.  
So from $d\mid a$ and $f\mid a$ you have $df\mid a$, and so $df\mid d'$. 
A: We do the part that you asked explicitly about, showing  that $df$ divides $d'$. 
Note that $d$ and $f$ are relatively prime. For if $w$ divides $d$ and $f$, then $w$ divides $b$ and $c$. But $b$ and $c$ are relatively prime, so $w$ divides $1$.
Now note that $d$ divides $d'$. For since $d$ is a common divisor of $a$ and $b$, it is a common divisor of $a$ and $bc$, so it divides the gcd of $a$ and $bc$. Similarly, $f$ divides $d'$. Since $d$ and $f$ are relatively prime, it follows that $df$ divides $d'$.
Remark: At the end we used a standard theorem, that if $u$ and $v$ are relatively prime and $u$ and $v$ divide $k$, then $uv$ divides $k$.
For a proof, we have $k=ut$ for some $s$, so $v$ divides $us$. By Euclid's Lemma, it follows that $v$ divides $s$, say $s=tv$, and therefore $k=uvt$.
