it is given that (a-b)x^2 + (b-c)x + (c – a) = 0
=> d=b^2 – 4ac
it is given that a = (a-b), b= (b-c), c = (c-a)
Now, d = (b -c)^2 – 4 * (a-b)(c-a) = 0
by using identity (a -b)^2 = a^2 + b^2 – 2*a*b
=> b^2 + c^2 -2bc – 4[ac – a^2 -bc + ab] =0
=> b^2 + c^2 -2bc – 4ac +4a^2 + 4 bc – 4ab =0
=> (2a^2) + (b^2) + (c^2) + 2(-2a)(c) + 2(b)(c) + 2(-2a)(b) =0
by using formula
=> a^2 + b^2 + c^2 + 2ab + 2bc +2ca = (a + b+ c)^2
=> -2a + b + c = 0
=> b + c = 2a (proved)
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