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