How do you show that: (a–b) (a+b) +(b–c) (b+c) +(c+a) (c–a) =0?

To show that (a-b)(a+b) + (b-c)(b+c) + (c+a)(c-a) = 0, we can start by expanding each of the terms:

(a-b)(a+b) = a^2 - b^2

(b-c)(b+c) = b^2 - c^2

(c+a)(c-a) = c^2 - a^2

Substituting these expressions back into the original equation, we get:

(a^2 - b^2) + (b^2 - c^2) + (c^2 - a^2) = 0

We can see that the terms cancel out in pairs, leaving us with 0 on the left-hand side. Therefore, the equation (a-b)(a+b) + (b-c)(b+c) + (c+a)(c-a) = 0 is true.