Prove that $\operatorname{lcm}(a,b,c)=\frac{abc \operatorname{gcd}(a,b,c)} {\operatorname{gcd}(a,b)\operatorname{gcd}(b,c)\operatorname{gcd}(c,a)}$

Asked Feb 20 '20 at 17:24

Active Feb 21 '20 at 06:39

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Prove that for any positive integers $a,b,c$ we have

$\operatorname{lcm}(a,b,c)=\dfrac{abc \operatorname{gcd}(a,b,c)} {\operatorname{gcd}(a,b)\operatorname{gcd}(b,c)\operatorname{gcd}(c,a)}$

To prove this, I used this: for two positive integers $a,b$ is $\operatorname{lcm}(a,b)=\frac{ab}{\gcd(a,b)}$

but I could do nothing, can you help?

edited Jun 12 '20 at 10:38

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asked Feb 20 '20 at 17:24

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But that Lemma is already linked in the first dupe. – Bill Dubuque Feb 20 '20 at 17:39
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We can also prove it using unique prime factorizations as here. Then it boils down to the following equality on exponents $\ \gamma = (\alpha+\beta+\gamma)+\alpha-(\alpha+\beta+\alpha).\ $ See also here. – Bill Dubuque Feb 21 '20 at 01:22
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It is similar to the Inclusion-Exclusion principle for three sets. – Geoffrey Trang Feb 21 '20 at 01:29
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@GeoffreyTrang Yes, inclusion-exclusion is one common way to prove identities between gcd and lcm. Some of these identities are already discussed here. – Bill Dubuque Feb 21 '20 at 01:48

Prove that $\operatorname{lcm}(a,b,c)=\frac{abc \operatorname{gcd}(a,b,c)} {\operatorname{gcd}(a,b)\operatorname{gcd}(b,c)\operatorname{gcd}(c,a)}$

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