How do you show that: (a–b) (a+b) +(b–c) (b+c) +(c+a) (c

To prove that (a-b)(a+b) + (b-c)(b+c) + (c+a)(c-a) = 0, we can expand each term and simplify:

(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) // using distributive property

= a^2 + ab - ab - b^2 + b^2 + bc - bc - c^2 + ac - ac - a^2

= 0 // combining like terms

Therefore, we have shown that (a-b)(a+b) + (b-c)(b+c) + (c+a)(c-a) = 0.

AnswerAi