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If loga/(b-c)=logb/(c-a)=logc/(a-b),then...

If `loga/(b-c)=logb/(c-a)=logc/(a-b)`,then` a^(b+c)+b^(c+a)+c^(a+b)` is

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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)`
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