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We have 2^a=3^b=6^c
                          = 2^c *3^c      --(1)
 This gives          2^c*3^c=3^b
                          ⇒    2^c= 3^(b-c)
  and        2^c*3^c= 2^a
               ⇒    3^c= 2^(a-c)
substituting value in one we have 2^c*3^c= 3^(b-c)*3^(a-c)
then we have  c= b-c
                     ⇒ b=2c
   and         c=a-c
              ⇒ a = 2c 
RHS= ab/(a+b)= 2c*2c/( 2c+2c)=c=LHS
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