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?
 A: It's much simpler using the identities you already know (associativity, commutativity, etc)
$$((a,b),(a,c))\, =\, ((a,b),a,c)\, =\, (a,b,a,c)\, =\, (a,b,c)$$
Remark $\ $ By induction, in the same way, one can always "flatten" such gcd expressions.
A: Your proof seems okay. This is probably the shortest one. But gonna be large because I am oversimplifying.
Let 
$$a=\prod_{i=1}^{k}p_i^{\alpha_i},b=\prod_{i=1}^{k}p_i^{\beta_i},c=\prod_{o=1}^{k}p_i^{\gamma_i}$$
Now $\displaystyle \text{gcd}(a,b,c)=\prod_{i=1}^{k}p_i^{\min(\alpha_i,\beta_i,\gamma_i)}$
and $\displaystyle \text{gcd}(a,b)=\prod_{i=1}^{k}p_i^{\min(\alpha_i,\beta_i)},\text{gcd}(a,c)=\prod_{i=1}^{k}p_i^{\min(\alpha_i,\gamma_i)}$
So prove the equality we need to prove the fact :
for arbitrary integers(also true for reals) $\alpha,\beta,\gamma$ 
$$\min(\alpha,\beta,\gamma)=\min (\min(\alpha,\beta),\min(\alpha,\gamma))$$
Which is obvious since we may WLOG assume $\alpha\ge \beta\ge \gamma$ and both sides equal $\gamma$. Hope this proof satisfies you.
A: Your proof seems OK. A more elegant proof would be to show that $m$ divides $a,b$ and $c$ if and only if $m$ divides $(a,b)$ and $(a,c)$, which is fairly routine. Then of course $(a,b,c)=((a,b),(a,c))$.
A: You can use just the one identity you know along with symmetry, and  nothing else, to simplify
$$ ((a,b),(a,c)) = ((a,b),a,c) = (((a,b),a), c) = ((a,a,b),c) = (((a,a),b),c) = ((a,a),b,c)$$
In fact, the associative identity for a binary operator $((a,b),c) = (a,(b,c))$ -- or in its more common expression for operators with infix notation $(a \cdot b) \cdot c = a \cdot(b\cdot c)$ -- is enough to prove that if you operate on an arbitrary number of things, it doesn't matter how you group them: e.g. if you know that binary gcds are associative, then it automatically follows, e.g., that
$$ ((a,(b,c)),d) = ((a,b),(c,d)) $$
without any fuss. And it means that extending the operator to any positive, finite number of terms is unambiguous: if I write
$$ (a,b,c,d) $$
then it doesn't matter how I group the terms into pairwise gcds, I have to get the same value.
If the operator is also symmetric (more commonly called "commutative" when talking about operations, but "symmetric" -- that is, $(a,b) = (b,a)$ or for infix operations $a\cdot b = b \cdot a$ -- that means you can rearrange the terms arbitrarily.
And thus, I immediately know that it makes sense to just say
$$ ((a,b),(a,c)) = (a,a,b,c) $$
and furthermore, if I instead write it as
$$ ((a,b),(a,c)) = ((a,a),(b,c))$$
this could be proven without ever resorting to ternary or quaternary gcds, just by using the associative and commutative laws.
A: 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.
