If (a + b + c)(a – b + c) = a2 + b2 + c2 , show that a, b, c are in

1.	If (a + b + c)(a – b + c) = a2 + b2 + c2 , show that a, b, c are in continued proportion.
Answer» (a + b + c)(a – b + c) = a² + b² + c² …[Given] ∴ a(a – b + c) + b(a – b + c) + c(a – b + c) = a² + b² + c² ∴ a² – ab + ac + ab – b² + be + ac – be + c² = a² + b² + c² ∴ a² + 2ac – b² + c² = a² + b² + c² ∴ 2ac – b² = b² ∴ 2ac = 2b² ∴ ac = b² ∴ b² = ac ∴ a, b, c are in continued proportion.

If (a + b + c)(a – b + c) = a2 + b2 + c2 , show that a, b, c are in continued proportion.

Answer»

(a + b + c)(a – b + c) = a² + b² + c² …[Given]

∴ a(a – b + c) + b(a – b + c) + c(a – b + c) = a² + b² + c²

∴ a² – ab + ac + ab – b² + be + ac – be + c² = a² + b² + c²

∴ a² + 2ac – b² + c² = a² + b² + c²

∴ 2ac – b² = b²

∴ 2ac = 2b²

∴ ac = b²

∴ b² = ac

∴ a, b, c are in continued proportion.