Question

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

Answer 1

Answer:

here is your answer.....

Step-by-step explanation:

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

a²-b²+b²-c²+c²-a²[all the terms are cancelled..]

=0 {proved}.

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Answer 2

Answer:

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

$(x - y)(x + y) = (x - y)^{2}$

So, similarly....

(a-b )(a+b)= (a-b) 2

(b-c) (B+C) = (b-c) 2

(C+a) (c-a) = (c-a) 2

$(a - b) ^{2} +( b - c) ^{2} + (c - a) ^{2}$

a2 - a2 + b2 - b2 + c2 - c2 = 0

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