Question

if a2/a+b+b2/b+c+c2+c+a=1 show that 1/a+b+1/b+c+1/c+a=0​

Answer 1

Answer:

1/a^2 + 1/b^2 + 1/c^2 = 1/ab + 1/bc + 1/ ac

=> 2/a^2 + 2/b^2 + 2/c^2 = 2/ab + 2/ bc + 2/ac

=>1/a^2 - 2/ab + 1/b^2 + 1/b^2 - 2/bc + 1/c^2 + 1/c^2 - 2/ac + 1/a^2 = 0

=> (1/a - 1/b)^2 + (1/b - 1/c)^2 + (1/c - 1/a)^2 = 0

=> 1/a - 1/b = 0, 1/b - 1/c = 0, 1/c - 1/a = 0 ( since sum of the perfect squares is 0)

=> 1/a = 1/b, 1/b = 1/c , 1/c = 1/a

=> a = b,. b = c, c = a => a= b = c

Step-by-step explanation:

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