If and a, b, c >0, then show that,

i. (a + b+ c)(b – c) = ab – c²

ii. (a² + b²)(b² + c²) = (ab + bc)²

iii. (a² + b²)/ab = (a + c)b

(i)

Given: a/b = b/c

⇒ b² = ac

Consider (a + b+ c)(b – c) = ab – ac + b² – bc + cb – c²

= ab – ac + ac – c² (∵ b² = ac)

= ab – c²

(ii)

Given:

a/b = b/c

⇒ b² = ac

Consider (a² + b²)(b² + c²) = a²b² + a² c² + b² b² + b²c²

= a²b² + ac(ac)+ b²(ac)+ b²c² (∵ b² = ac)

= a²b² + b²(ac)+ b²(ac)+ b²c² (∵ b² = ac)

= a²b² + 2b²(ac)+ b²c²

= a²b² + 2ab²c+ b²c²

= (ab + bc)²

(iii)

Given: a/b = b/c

⇒ b² = ac

Consider (a² + b²)/ab = (a² + ac)/ab (∵ b² = ac)

= (a + c)/b