Question

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

Answer 1

L.H.S = (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a)

Using identity:

(a – b) (a + b) = a2 – b2

We get,

= (a2 – b2) + (b2 – c2) + (c2 – a2)

= a2 – b2 + b2 – c2 + c2 – a2

= 0

= R.H.S

Hence, verified

hope this helps you.....

Answer 2

Answer:

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

=a²-b²+b²-c²+c²-a²

=0