Question

show that
(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)=0
​

Answer 1

Answer:

$using \: identity \\ {x}^{2} - {y}^{2} = (x + y)(x - y)$

${a}^{2} - {b}^{2} + {b}^{2} - {c}^{2} + {c}^{2} - {a}^{2} \\ = 0$

Answer 2

Step-by-step explanation:

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

=> a^2-b^2+b^2-c^2+c^2-a^2

=> 0