Show that $abc=[ab,bc,ca]*(a,b,c)=(ab,bc,ca)*[a,b,c]$ Let $a,b \in \mathbb N$. Show that
$$ abc=[ab,bc,ca](a,b,c)=(ab,bc,ca)[a,b,c] .$$
How would I prove this?

If $\gcd(a,b,c) = d > 1$, then by writing:  $a = pd$, $b = qd$, and $c = rd$, we can factor out $d^3$ in common. So we can assume $\gcd(a,b,c) = 1$. Then we need to prove: $abc = [ab,bc,ca]$. Let $[ab,bc,ca] = m$, then: $ab|m$, $bc|m$, and $ca|m$. Thus:
$abc|mc$, $abc|ma$, and $abc|mb$. Thus:
$abc|\gcd(ma,mb,mc) = m\cdot \gcd(a,b,c) = m\cdot 1 = m$. Thus: $abc|m$. We only need to prove: $m|abc$.
But observe that: $ab|abc$, $bc|abc$, and $ca|abc$. Thus: $[ab,bc,ca]|abc$. This means: $m|abc$.
Thus: $abc|m$, and $m|abc$. So: $m = abc$. Done.
