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

If (log a)/(b-c)=(log b)/(c-a)=(log c)/(a-b), then a^(b+c)*b^(c+a)*c^(a+b)=

If (log a)/(b-c)=(log b)/(c-a)=(log c)/(a-b), then a^(b+c)+b^(c+a)+c^(a+b) is

If (loga)/(b-c) = (logb)/(c-a) = (logc)/(a-b) , then prove that a^(a)b^(b)c^(c)=1 .

If (a(b + c - a))/(log a) = (b (c + a - b))/(log b) = (c (a + b - c))/(log c ) then (a^(b).b^(a))/(c^(a).a^(c)) equals