Math, asked by anupam6548, 12 hours ago

if a/b=b/c then prove that (a+b+c) (a-b+c) =a^2+b^2+c^2​

Answers

Answered by Anonymous
2

Consider, a

2+b 2+c 2–ab–bc–ca=0

Multiply both sides with 2, we get

2(a 2+b 2+c 2

–ab–bc–ca)=0

⇒ 2a 2+2b 2+2c 2

–2ab–2bc–2ca=0

⇒ (a 2–2ab+b 2)+(b 2–2bc+c 2)+–2c )=0

⇒ (a–b) 2

+(b–c) 2

+(c–a) 2

=0

Since the sum of square is zero then each term should be zero

⇒ (a–b) 2

=0,(b–c) 2

=0,(c–a) 2

=0

⇒ (a–b)=0,(b–c)=0,(c–a)=0

⇒ a=b,b=c,c=a

∴ a=b=c.

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