Question

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

Answer 1

as we know the identity (a-b)(a+b) = a²-b²
also (b-c)(b+c) = b²-c²
and (c-a)(c+a)=c²-a²
Therefore, Substituing this in question

(a²-b²)+(b²-c²)+(c²-a²)

By opening brackets
a²-b²+b²-c²+c²-a²

a²-a²+b²-b²+c²-c²

0+0+0 = 0

So answer is 0

Answer 2

To prove;

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

Proof;

L.H.S.

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

=> {(a-b)(a+b)} + {(b+c)(b-c)} + {(c-a)(c+a)}

According to algebric identities one of the identity is that :-

=> (x-y)(x+y) = x² - y²

So,

(a-b) (a+b) = a² - b²

(b-c)(b+c) = b² - c²

(c-a)(c+a) = c² - a²

So, putting values ;

=> a² - b² + b² - c² + c² - a²

=> a² - a² + b² - b² + c² - c²

=> 0 [<---- L.H.S.]

R.H.S.

=> 0 [<----- R.H.S.]

• Since, L.H.S. = R.H.S.

Therefore, proved that (a-b) (a+b) + (b-c) (b+ c) + (c-a) (c + a) = 0​