Question

(b+c)(b-c)+(c+a)(c-a)+(a+b)(a-b)​

Answer 1

Answer:

Step-by-step explanation:

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

, =(a²-b²)+(b²-c²)+(c²-a²),

=(a²-b²+b²-c²+c²-a²),

In this case, each variable cancels it's negative version. = 0

Answer 2

Answer:

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

=(a2-b2)+(b2-c2)+(c2-a2)

=(a2-b2+b2-c2+c2-a2)

=0