Question

prove that (a-b)(a+b)+(b-c)(b+c)=a²-c²​

Answer 1

Consider, a2+b2+c2–ab–bc–ca=0

Multiply both sides with 2, we get

2(a2+b2+c2–ab–bc–ca)=0

⇒ 2a2+2b2+2c2–2ab–2bc–2ca=0

⇒ (a2–2ab+b2)+(b2–2bc+c2)+(c2–2ca+a2)=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.

Answer 2

Hope the answer is helpful for you...