Question

Show that
(a-b) (a+b) + (b-c) (b+c) + (c-a) (c+a) = 0.​

Answer 1

Step-by-step explanation:

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

from L.H.S:

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

=a²-b²+b²-c²+c²-a²

=a²-a²-b²+b²-c²+c²

=0

hence proved

Answer 2

Answer:

From LHS.

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

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

=a²+ab-ab-b²+b²+bc-bc-c²+c²+ac-ac-a².

=0.

Therefore, LHS=RHS.