Prove $\gcd(a,b) \gcd(a,c) \gcd(b,c) \,\text{lcm} (a,b,c)^2=$ $\text{lcm}(a,b)\,\text{lcm}(a,c) \,\text{lcm}(b,c) \gcd(a,b,c)^2$ [closed]

By here $\ [a,b,c]\, =\, \dfrac{abc}{(ab,bc,ca)} =\dfrac{abc(a,b,c)}{(a,b)(b,c)(c,a)} $ by $\ (a,b)(b,c)(c,a) \overset{\rm\color{#c00}E}= (a,b,c)(ab,bc,ca) $

Squaring that yields your equation (after replacing $ab/(a,b)$ by $[a,b]$ etc).

Or let $\ell = {\rm lcm},\ g = {\rm gcd}.\,$ Using $\,\color{#c00}{\ell}(x,y) = xy/g(x,y)$ to replace red lcms by gcds

$$\color{#c00}\ell(a,\ell(b,c))= \dfrac{a\,\color{#c00}\ell(b,c)}{\underbrace{g(a,\ell(b,c))}_{\large \!\!\!\color{#c00}\ell(g(a,b),g(a,c))\!\!\!}} = \dfrac{a\, bc/g(b,c)}{g(a,b)g(a,c)/g(a,b,c)} = \dfrac{abc\,g(a,b,c)}{g(a,b)g(b,c)g(c,a)}\qquad$$

where we used gcd distributes over lcm in underbrace on bottom of the first fraction, and in the bottom of the second we use $\,g(g(a,b),g(a,c)) = g(a,b,c)\,$ by gcd is associative. The dual identity follows by swapping gcd $\leftrightarrow$ lcm.

Remark $ $ The final equality $(\overset{\rm\color{#c00}E}=)$ in the first line follows simply by gcd "polynomial" arithmetic, i.e. by applying the gcd associative, commutative and distributive laws to expand the products on both sides, which yields that both $= (aab,aac,abb,abc,acc,bbc,bcc),\,$ as here.

Let's write $$a=p_1^{a_1}\cdots p_r^{a_r},$$ $$b=p_1^{b_1}\cdots p_r^{b_r},$$ $$c=p_1^{c_1}\cdots p_r^{c_r},$$ where $p_i$ are primes. Now

$$\displaystyle \rm{gdc}(a,b)gcd(a,c)gcd(b,c)lmc(a,b,c)^2=\Pi_{i=1}^rp_i^{\min\{a_i,b_i\}+\min\{a_i,c_i\}+\min\{b_i,c_i\}+2\max\{a_i,b_i,c_i\}}$$ and

$$\displaystyle \rm{lcm}(a,b)lcm(a,c)lcm(b,c)gcd(a,b,c)^2=\Pi_{i=1}^rp_i^{\max\{a_i,b_i\}+\max\{a_i,c_i\}+\max\{b_i,c_i\}+2\min\{a_i,b_i,c_i\}}$$

Just check that $$\min\{a_i,b_i\}+\min\{a_i,c_i\}+\min\{b_i,c_i\}+2\max\{a_i,b_i,c_i\}=\max\{a_i,b_i\}+\max\{a_i,c_i\}+\max\{b_i,c_i\}+2\min\{a_i,b_i,c_i\}.$$ WLOG assume $a_i\le b_i\le c_i$ and see that both terms are $2a_i+b_i+2c_i.$