Question

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

Answer 1

Step-by-step explanation:

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

a^2+ab_ab-b^2+b^2+bc-bc-c^2+c^2+ac-ac-a^2=0

0=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​