Is (a-b) (a+b) + (b-c)(b+c) + (c-a)(c+a) = 0 , Yes or no ( reason not needed )
Answers
Answered by
1
Given =>
(a - b)(a + b) + (b - c)(b + c) + (c - a)(c + a) = 0
Solving LHS
=> (a - b)(a + b) + (b - c)(b + c) + (c - a)(c + a)
(using formula (a - b)(a + b) = a² - b²)
=> a² - b² + b² - c² + c² - a²
=> 0
=> RHS
Therefore,
LHS = RHS
Hence Proved (Yes).
(a - b)(a + b) + (b - c)(b + c) + (c - a)(c + a) = 0
Solving LHS
=> (a - b)(a + b) + (b - c)(b + c) + (c - a)(c + a)
(using formula (a - b)(a + b) = a² - b²)
=> a² - b² + b² - c² + c² - a²
=> 0
=> RHS
Therefore,
LHS = RHS
Hence Proved (Yes).
Tomboyish44:
Thank you !!!!
Answered by
1
Yes....it is correct...
(a+b)(a-b) + (b+c)(b-c) + (c+a)(c-a)
= a²-b² + b²-c² + c²- a
= 0
(a+b)(a-b) + (b+c)(b-c) + (c+a)(c-a)
= a²-b² + b²-c² + c²- a
= 0
Similar questions