Let `loga/(b-c) = logb/(c-a) = logc/(a-b) = k->(1)`
Then, `loga+logb+logc = k(b-c+c-a+a-b)`
`=>log(abc) = 0`
`=>log(abc) = log1`
`=>abc = 1->(2)`
Now, from (1),
`(aloga)/(a(b-c)) = (blogb)/(b(c-a)) =(c logc)/(c(a-b)) = k`
`=>aloga+blogb+clogc = k(ab-ac+bc-ab+ca-cb)`
`=>loga^a+logb^b+logc^c = 0`
`=>loga^a+logb^b+logc^c = log1`
`=>a^ab^bc^c = 1->(3)`
Now, `a^(b+c)b^(c+a)c^(a+b) = (a^(a+b+c)b^(a+b+c)c^(a+b+c))/(a^ab^bc^c) `
`= (abc^(a+b+c))/(abc)^(abc)`
From (2) and (3),
`a^(b+c)b^(c+a)c^(a+b) = 1/1 = 1`
