Show that (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) =

1.	Show that (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0
Answer» Step-by-step explanation: to prove : (a – B) (a + b) + (b – C) (b + c) + (c – a) (c + a) = 0 take LHS, = (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = a²-b² + b²-c² + c²-a² = a²--a² -b² + b² -c² + c² = 0 = RHS => LHS = RHS Hence proved

Show that (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0

Answer»

Step-by-step explanation:

to prove : (a – B) (a + b) + (b – C) (b + c) + (c – a) (c + a) = 0

take LHS,

= (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a)

= a²-b² + b²-c² + c²-a²

= a²--a² -b² + b² -c² + c²

= 0

= RHS

=> LHS = RHS

Hence proved

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Show that (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0​

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Show that (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = 0