How to show that $\displaystyle [a,b,c] = \frac{abc}{(ab,bc,ca)}$ without prime factorization? I think this has been asked before, but I couldn't find it on math.SE. I googled it too, but I wasn't lucky enough to find it there either. So, here's the problem:
Demonstrate that for any $a,b,c \in \mathbb{N}$: $$\displaystyle [a,b,c] = \frac{abc}{(ab,bc,ca)}$$
It's not very hard to verify this equality once we know the prime factorization theorem. We can set:
$$a = p_1^{\alpha_1} p_2^{\alpha_2}\cdots p_n^{\alpha_n}$$
$$b = p_1^{\beta_1} p_2^{\beta_2} \cdots p_n^{\beta_n}$$
$$c = p_1^{\gamma_1} p_2^{\gamma_2}\cdots p_n^{\gamma_n}$$
Where $0 \leq \alpha_i,\beta_i,\gamma_i $ and $p_i$ is a prime number that appears in the prime factorization of at least one of the three.
Therefore, the equality is true if and only if this equality holds for all $1 \leq i \leq n$:
$$\min\{\alpha_i+\beta_i,\beta_i+\gamma_i,\gamma_i+\alpha_i\} + \max\{\alpha_i,\beta_i,\gamma_i\} = \alpha_i+\beta_i+\gamma_i$$
It's very easy to verify this equality. Therefore, the theorem holds if we are allowed to use prime factorization theorem.
How can we demonstrate that the equality is true without using the prime factorization theorem?
 A: A proof is mechanical using universal properties $\rm\,\color{#c00}{(UP)\ of\ lcm,gcd}\,$ and $\,\rm\color{#0a0}{duality}$
Theorem $\,\ {\rm lcm}(a,b,c)\, =\, \dfrac{abc}{(bc,ca,ab)}$
$\begin{eqnarray}{\bf Proof}\quad\ \ \ {\rm lcm}(a,b,c)&\mid&\ n\\ 
\iff a,b,c&\mid&\ n,\ \ \text{by}\,\color{#c00}{\text{  UP lcm}}\\ 
\iff\quad abc&\mid&\, nbc,nca,nab,\ \ \ \ \text{by}\ \ \rm\color{#0a0}{duality},\ see\ below\\ 
\iff\quad abc&\mid&\! (nbc,nca,nab),\ \ \text{by}\,\color{#c00}{\text{  UP gcd}}\\
\iff\quad abc&\mid&\, n(bc,ca,ab) \\ 
\iff\ \ \dfrac{abc}{(bc,ca,ab)}&\Big|&\ n\end{eqnarray}$
Remark $\ $ The innate symmetry governing the proof is the involution $\ x\to x' = abc/x\,$ which highlights the duality $ $ lcm$' =$  gcd,$\ $ gcd$' =$ lcm, $ $ that arises from $\, \color{#0a0}{x\mid y\iff y'\mid x'},\,$ viz.
$\qquad\qquad \begin{eqnarray} {\rm lcm}(a,b,c)\! &=&\, {\rm lcm}(a,b,c)''\\
&=&\, {\gcd(a',b',c')'}\\
&=&\,  \dfrac{abc}{\gcd(a',b',c')}\\
&=&\, \dfrac{abc}{\gcd(bc,ca,ab)}\end{eqnarray}$
A: Notice that we can wirte $[a,b,c]=[[a,b],c]$ and let say $k=[a,b]=\dfrac {ab}{(a,b)}$
$[k,c]=\dfrac {kc}{(k,c)}=\dfrac {abc}{(a,b)(k,c)}$ and notice that $r(x,y)=(rx,ry)$
$$=\dfrac {abc}{((a,b)k,(a,b)c)}$$
$$=\dfrac {abc}{(ab,(a,b)c)} $$
$$=\dfrac {abc}{(ab,(ac,bc))}$$
$$=\dfrac {abc}{(ab,ac,bc)}$$
Note: $(a,b),[a,b]$ stand for $\gcd(a,b)$ and $\text{lcm}[a,b]$ respectively.
