GCD, LCM Relationship

Question

Problem: Let $(a,b)$ denote the greatest common divisor of $a$ and $b$ and $[a,b]$ denote the least common multiple of $a$ and $b$. Similarly for three integers.

Prove that \begin{align} \frac{[a,b,c]^2}{[a,b][b,c][c,a]}=\frac{(a,b,c)^2}{(a,b)(b,c)(c,a)} . \end{align}

I see that this equality holds for all positive integers, however, how would I actually prove it?

Answer 1

Using the properties: $$(a,b,c)=((a,b),c)=(a,(b,c))=((a,c),b) \\ [a,b,c]=[[a,b],c]=[a,[b,c]]=[[a,c],b] \\ (a,b) \cdot [a,b]=a \cdot b \Rightarrow [a,b]=\frac{a \cdot b}{(a,b)}$$ we get:

$$\frac{[a,b,c]^2}{[a,b][b,c],[c,a]}=\frac{\frac{a^2 b^2 c^2}{(a,b,c)^2}}{\frac{ab}{(a,b)}\frac{bc}{(b,c)},\frac{ca}{(c,a)}}=\frac{a^2b^2c^2(a,b)(b,c)(c,a)}{(a,b,c)^2abbcca}=\frac{(a,b)(b,c)(c,a)}{(a,b,c)^2}$$

Answer 2

Suppose that $a=p^\alpha, b=p^\beta, c=p^\gamma$. Because both expressions are symmetric, assume that $\alpha\le \beta \le \gamma$. (otherwise there are cases to consider)

Then the LHS is $p^k$, where $$k=2\max(\alpha,\beta,\gamma)-\max(\alpha,\beta)-\max(\beta,\gamma)-\max(\gamma,\alpha)=2\gamma-\beta-\gamma-\gamma=-\beta$$

On the other hand, the RHS is $p^m$, where $$m=2\min(\alpha,\beta,\gamma)-\min(\alpha,\beta)-\min(\beta,\gamma)-\min(\gamma,\alpha)=2\alpha-\alpha-\beta-\alpha=-\beta$$

Hence the LHS and RHS have the same power of $p$, namely $-\beta$. Since GCD and LCM are multiplicative functions, in fact LHS=RHS for all $a,b,c$.