Prove $(a,b,c)=((a,b),(a,c))$ The notation is for the greatest common divisor.  I know that 
$$(a,b,c)=((a,b),c)=((a,c),b)=(a,(b,c))$$
Suppose $g=(a,b,c)$.  Then $g\mid a,b,c$.  Also, $g\mid(a,b),c$ and $g\mid(a,c),b$.  Thus there exist integers $k,m$ such that
$$(a,b)=gk, (a,c)=gm$$
Then
$$((a,b),(a,c))=(gk,gm)=g(k,m)$$
Therefore, since $(k,m)=1$ (otherwise $g$ would not be the greatest common divisor of $(a,b)$ and $(a,c)$), $g=((a,b),(a,c))$.
Is this proof okay?

It's worthwhile to consider a proof that resides entirely within the algebra of divisibility.
The main trick in this regard is that if $x$ and $y$ are nonnegative numbers and you can show $x \mid y$ and $y \mid x$, then it follows that $x = y$
Thus, to prove $((a,b),(a,c)) = (a,b,c)$, you need to prove


*

*$((a,b),(a,c)) \mid (a,b,c)$

*$(a,b,c) \mid ((a,b),(a,c)) $


To get you started on both points, observe that the definition of gcd means that the second statement is equivalent to


*

*$(a,b,c) \mid (a,b)$ and $(a,b,c) \mid (a,c) $


The entire proof ultimately boils down to repeatedly writing down statements equivalent to what you're trying to prove until you're left with something obviously true.
