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» LHS (a – b) (a + b) + (b – c) (b + c) + (c – a) (c + a) = (a² – b²) + (b² – c²) + (c² – a²) = a² – b² + b² – c² + c² – a² = (a² – a²) + (b² – b²) + (c² – c²) = 0 + 0 + 0 + 0 = RHS

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

Answer»

LHS

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

= (a² – b²) + (b² – c²) + (c² – a²)

= a² – b² + b² – c² + c² – a²

= (a² – a²) + (b² – b²) + (c² – c²)

= 0 + 0 + 0 + 0

= RHS