(i) a(b² + c²) = c(a² + b²)

Given that a, b, c are in GP.

By using the property of geometric mean,

b² = ac

Let us consider LHS: a(b² + c²)

Now, substituting b² = ac, we get

a(ac + c²)

a²c + ac²

c(a² + ac)

Substitute ac = b² we get,

c(a² + b²) = RHS

∴ LHS = RHS

Hence proved.

(ii) a²b²c² [1/a³ + 1/b³ + 1/c³] = a³ + b³ + c³

Given that a, b, c are in GP.

By using the property of geometric mean,

b² = ac

Let us consider LHS: a²b²c² [1/a³ + 1/b³ + 1/c³]

a²b²c²/a³ + a²b²c²/b³ + a²b²c²/c³

b²c²/a + a²c²/b + a²b²/c

(ac)c²/a + (b²)²/b + a²(ac)/c [by substituting the b² = ac]

ac³/a + b⁴/b + a³c/c

c³ + b³ + a³ = RHS

∴ LHS = RHS

Hence proved.

(iii) (a + b + c)² / (a² + b² + c²) = (a + b + c) / (a - b + c)

Given that a, b, c are in GP.

By using the property of geometric mean,

b² = ac

Let us consider LHS: (a + b + c)² / (a² + b² + c²)

(a + b + c)² / (a² + b² + c²) = (a + b + c)² / (a² – b² + c² + 2b²)

= (a + b + c)² / (a² – b² + c² + 2ac) [Since, b² = ac]

= (a + b + c)² / (a + b + c)(a - b + c) [Since, (a + b + c)(a - b + c) = a² – b² + c² + 2ac]

= (a + b + c) / (a - b + c)

= RHS

∴ LHS = RHS

Hence proved.

(iv) 1/(a² – b²) + 1/b² = 1/(b² – c²)

Given that a, b, c are in GP.

By using the property of geometric mean,

b² = ac

Let us consider LHS: 1/(a² – b²) + 1/b²

Let us take LCM

1/(a² – b²) + 1/b² = (b² + a² – b²)/(a² – b²)b²

= a² / (a²b² – b⁴)

= a² / (a²b² – (b²)²)

= a² / (a²b² – (ac)²) [Since, b² = ac]

= a² / (a²b² – a²c²)

= a² / a²(b² – c²)

= 1/ (b² – c²)

= RHS

∴ LHS = RHS

Hence proved.

(v) (a + 2b + 2c) (a – 2b + 2c) = a² + 4c²

Given that a, b, c are in GP.

By using the property of geometric mean,

b² = ac

Let us consider LHS: (a + 2b + 2c) (a – 2b + 2c)

Upon expansion we get,

(a + 2b + 2c) (a – 2b + 2c) = a² – 2ab + 2ac + 2ab – 4b² + 4bc + 2ac – 4bc + 4c²

= a² + 4ac – 4b² + 4c²

= a² + 4ac – 4(ac) + 4c² [Since, b² = ac]

= a² + 4c²

= RHS

∴ LHS = RHS

Hence proved.