Identity involving LCM and GCD: $\frac{[a,b,c]^2}{[a,b][b,c][c,a]}=\frac{(a,b,c)^2}{(a,b)(b,c)(c,a)}$

Let $(a,b)$ denote the GCD of $a$ and $b$, and let $[a,b]$ denote the LCM of $a$ and $b$. Prove $\frac{[a,b,c]^2}{[a,b][b,c][c,a]}=\frac{(a,b,c)^2}{(a,b)(b,c)(c,a)}$.

Solution 1:

Hint $\ $ Let $\,\alpha,\beta,\gamma $ be the powers of the prime $\,p\,$ in the unique prime factorization of $\,a,b,c,\,$ resp. $\ $ By symmetry we may choose notation so that $\,\alpha\le \beta\le \gamma.\,$ Comparing powers of $p$ on both sides below

$$\begin{align}\frac{[a,b,\color{#90f}c]^{\large\color{#90f}2}}{[a,\color{#0a0}b][b,\color{#c00}c][\color{darkorange}c,a]} &\,=\, \frac{(a,b,c)^{\large 2}}{(a,b)(b,c)(c,a)}\\[.4em] \iff\ \ \color{#90f}{2\gamma} - (\color{#0a0}\beta + \color{#c00}\gamma + \color{darkorange}\gamma) &\,=\, 2\alpha - (\alpha + \beta + \alpha)\end{align}$$

Alternatively $ $ combine $\ (ab,bc,ca)[a,b,c]\, =\, abc\, =\, [ab,bc,ca](a,b,c)\ $ from this answer with

$$\qquad (a,b,c)(bc,ca,ab) = (a,b)(b,c)(c,a)$$

which is true since both are equal to $\ (aab,aac,abb,abc,acc,bbc,bcc)\ $ by the distributive law.