Formulas for lcm of three integers

Question

I know that both of the following formulas equal to $\text{lcm}(a,b,c)$, but how can we show that

$\dfrac{abc \cdot \gcd(a,b,c)}{\gcd(a,b) \cdot \gcd(a,c) \cdot \gcd(b,c)} = \dfrac{abc }{\gcd(ab,ac,bc)}$ ?

Answer 1

As here, cancelling $\,abc\,$ then clearing denominators shows it is equivalent to

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

This equality is easy by gcd "polynomial" arithmetic: expanding the products on both sides (apply gcd associative, commutative, distributive laws), shows both $= (aab,aac,abb,abc,acc,bbc,bcc)$.

Note: as shown here, it is equivalent to gcd distributes over lcm (or vice-versa).

Answer 2

Hint:

WLOG the highest exponents of prime $p$ that divides $a,b,c$ be $A,B,C$

WLOG $A\ge B\ge C\ge0$

So, the highest exponents of prime $p$ that divides the numerator of the LHS will be $$A+B+C+C$$

For the denominator of the left hand side the highest exponent $=B+C+C$

For the denominator of the right hand side the highest exponent $$=B+C$$

The above will hold true for any prime that divides $abc$