Question

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

Answer 1

Answer:

The answer is a^2 - b^2+b^2-c^2 + c^2- a^2 = 0

Step-by-step explanation:

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

= (a^2+ab-ab -b^2) + ( b^2+ bc - bc -c^2 ) + (c^2+ca-ca -a^2)

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

= 0

( Proved )

Answer 2

Answer:

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

(a²-b²) + (b²-c²) + (c²-a²) = 0

a² - b² + b² - c² + c² - a² = 0

0 = 0 (all variable will be cancelled by each other)

I hope it will help you

please mark me brainlist