Question

show that
\[\large \dfrac{(a,b,c)^2}{(a,b)(b,c)(c,a)}=\dfrac{[a,b,c]^2}{[a,b][b,c][c,a]}\]

where \((a,b,c)=\gcd(a,b,c)\)
\([a,b,c]=\text{lcm}(a,b,c)\)

\(a,b,c\in \mathbb{Z^{+}}\)

Answer 1

m tryin hard but i haven't learnt such things..is this question related to number theory?

Answer 2

ive got this instead :\
\(\Large \large \dfrac{(a,b,c)^2}{(a,b)(b,c)(c,a)}=\dfrac{[a,b][b,c][c,a]}{[a,b,c]^2 } \)

Answer 3

Let the prime factorizations of a, b, c be as follows:
\(p_1^{a_1}...p_k^{a_k}\)
\(p_1^{b_1}...p_k^{b_k}\)
\(p_1^{c_1}...p_k^{c_k}\)

Note that when we write it in this form some of the exponents might be zero since not all the primes have to be factors. But also note that at least one of \(a_i , b_i , c_i \) is non-zero because otherwise we wouldn't consider that prime at all.

Now, note that \((a,b) = p_1^{min(a_1,b_1)}...p_k^{min(a_k,b_k)}\) and \((a,b,c) = p_1^{min(a_1,b_1,c_1)}...p_k^{min(a_k,b_k.c_k)}\)

Similarly \( [a,b] = p_1^{max(a_1,b_1)}...p_k^{max(a_k,b_k)}\) and \([a,b,c] = p_1^{max(a_1,b_1,c_1)}...p_k^{max(a_k,b_k.c_k)}\)

Using these, the equality can be reduced to
\( 2min(a_i,b_i,c_i) - min(a_i,b_i) - min(b_i,c_i) - min(c_i,a_i)\)
\( = 2max(a_i,b_i,c_i) - max(a_i,b_i) - max(b_i,c_i) - max(c_i,a_i)\) for each i

That in turn can be reduced to
\(max(a_i,b_i) - min(a_i,b_i) + max(b_i,c_i) - min(b_i,c_i) + max(c_i,a_i) - min(c_i,a_i) \)
\(= 2[max(a_i,b_i,c_i) - min(a_i,b_i,c_i)]\)

that is, \( |a_i - b_i| + |b_i-c_i| + |c_i-a_i| = 2[max(a_i,b_i,c_i) - min(a_i,b_i,c_i)]\) which is easily verified

For example, WLOG assume \(c_i\) is the middle one, then,
\(|a_i - b_i| + |b_i-c_i| + |c_i-a_i| = 2|a_i-b_i|\)
[\(c_i\) cancels out since in the sum it occurs twice with opposite signs]

\(= 2[max(a_i,b_i) - min(a_i,b_i)]\)

\(= 2[max(a_i,b_i,c_i) - min(a_i,b_i,c_i)]\)

Thus for each i, we have
\[ \frac{p_i^{2min(a_i,b_i,c_i)}}{p_i^{min(a_i,b_i)}p_i^{min(b_i,c_i)}p_i^{min(c_i,a_i)}} = \frac{p_i^{2max(a_i,b_i,c_i)}}{p_i^{max(a_i,b_i)}p_i^{max(b_i,c_i)}p_i^{max(c_i,a_i)}}\]
Now multiplying both sides over \(i\) we get the desired equality.

Answer 4

i have reach this and went into a cycle :-\

Answer 5

I do not see how the second part be 1. Pairwise GCD of a,b,c will not be 1 unless they are coprime. 1/LCM is certainly smaller than 1 as LCM>=1 unless a,b,c=1.

Answer 6

that was pointless i'll try something else :)