Math, asked by javaidbhat15252, 5 months ago

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

Answers

Answered by mathdude500
0

Answer:

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

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

Answered by AbhaySachan
0

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

Similar questions