Now, as a,b,c are in GP.

Using the idea of geometric mean we can write –

∴ b² = ac …(1)

Put in the LHS of the given equation to be proved –

LHS = a(ac + c²) {putting b^{2 =} ac}

⇒ LHS = a²c + ac²

⇒ LHS = c(a² + ac)

Again put ac = b²

⇒ LHS = c(a² + b²) = RHS

∴ L.H.S = R.H.S

Hence proved