GCD, LCM Relationship 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? 
 A: 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 $a,b,c$.
A: 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: 
$$\frac{[a,b,c]^2}{[a,b][b,c],[c,a]}=\frac{\frac{a^2 b^2 c^2}{(a,b,c)^2}}{\frac{ab}{(a,b)}\frac{bc}{(b,c)},\frac{ca}{(c,a)}}=\frac{a^2b^2c^2(a,b)(b,c)(c,a)}{(a,b,c)^2abbcca}=\frac{(a,b)(b,c)(c,a)}{(a,b,c)^2}$$
A: Hint: 
For $p$ prime and $n \in \mathbb Z$, let $v_p(n)$ be the exponent of $p$ in the prime factorization of $n$.
Let $A = v_p(a)$,  $B= v_p(b)$, $C= v_p(c)$. Then we want
$$
\begin{align}
& \ 2\max(A,B,C) - \max(A,B) - \max(B,C) - \max(C,A)
\\
=& \ 2\min(A,B,C) - \min(A,B) - \min(B,C) - \min(C,A)
\end{align}
$$
which is quite easy to prove, assuming wlog that $A \le B \le C$.
