Math, asked by priya212, 1 year ago

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

durgeshg2k3: 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.

Answers

Answered by sneha3946

(a-b)(a+b)+b-c)(b+c)+(c-a)(c+a)
=a^2-b^2+b^2-c^2+c^2-a^2=0
a^2-c^2+c^2-a2=0. simplify
a^2-a^2=0. simplify
0=0

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