Question

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Answer 1

Answer:

Here,

(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0

when we solved it then we get

3x^2-2x(a+b+c)+(ab+bc+ac)

Let D be the discriminant

D=4(a+b+c)^2-12(ab+bc+ac)

=4[(a+b+c)^2-3(ab+bc+ac)]

=2[2a^2+2b^2+2c^2-2ab-2bc-2ac]

=2[(a-b)^2+(b-c)^2+(c-a)^2]

If D=0 then

(a-b)^2+(b-c)^2(c-a)^2=0

taking square root on both side

a-b+b-c+c-a=0

a-b=0,b-c=0,c-a=0

a=b, b=c, c=a

a=b=c

hence proved

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