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When I was learning about GCD and LCM I found this relation somewhere. But I didn't get proof for this relation. $$\frac{\gcd(a,b,c)^2}{\gcd(a,b)*\gcd(b,c)*\gcd(c,a)}=\frac{\operatorname{lcm}(a,b,c)^2}{\operatorname{lcm}(a,b)*\operatorname{lcm}(b,c)*\operatorname{lcm}(c,a)} $$ where GCD represents Greatest Common Divisor and LCM represents lowest common multiple.

Dev San
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    Write all the multiples and divisors using the primes that occur in the prime factorizations of $a$, $b$ and $c$. (This will look ugly in complete generality. If you write it out for, say $6, 15, 24$ you should see the argument.) – Ethan Bolker Dec 26 '18 at 16:28
  • There are many similar question at this site, where you can see how the argument goes, e.g., here. Have a look yourself at further (more similar) links. – Dietrich Burde Dec 26 '18 at 16:31

2 Answers2

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Let $a=\prod{p_i^{\alpha_i}}, b=\prod{p_i^{\beta_i}}$ and $c=\prod{p_i^{\gamma_i}}$ where $p_i$ denote the prime factors of a, b, c

We know that $\gcd(a,b)=\prod{p_i^{\max(\alpha_i,\beta_i)}}$, $\operatorname{lcm}(a,b)=\prod{p_i^{\min(\alpha_i,\beta_i)}}$

Therefore we have to show that $$2\max(\alpha_i,\beta_i,\gamma_i)-\max(\alpha_i,\beta_i)-\max(\beta_i,\gamma_i)-\max(\gamma_i,\alpha_i)=2\min(\alpha_i,\beta_i,\gamma_i)-\min(\alpha_i,\beta_i)-\min(\beta_i,\gamma_i)-\min(\gamma_i,\alpha_i)$$ for each index i.

Without loss of generality we can assume $\alpha_i\geq\beta_i\geq\gamma_i $ for any particular index i. Then above equation can reduce to $$2\alpha_i-\alpha_i-\beta_i-\alpha_i=2\gamma_i-\beta_i-\gamma_i-\gamma_i $$ which is the identity.

Sai Satwik Kuppili
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Suppose $a=dxza',b=dxyb',c=dyzc'$ and those numbers are realtive prime so $GCD(a,b,c)=d,GCD(a,b)=dx,GCD(b,c)=dy,GCD(c,a)=dz,LCM(a,b,c)=dxyza'b'c',LCM(a,b)=dxyza'b',LCM(b,c)=dxyzb'c',LCM(c,a)=dxyza'c'$ $$\frac{GCD(a,b,c)^2}{GCD(a,b)GCD(b,c)GCD(c,a)}=\frac{d^2}{dx\cdot dy\cdot dz}=\frac{1}{dxyz}$$ $$\frac{LCM(a,b,c)}{LCM(a,b)LCM(b,c)LCM(c,a)}=\frac{d^2x^2y^2z^2a'^2b'^2c'^2}{dxyza'b'\cdot dxyzb'c'\cdot dxyza'c'}=\frac{1}{dxyz}$$

Reynan Henry
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