Prove that 3sin⁻¹x = sin⁻¹(3x - x³), x ∈ [-1/2, 1/2]

Let x = sin θ. Since x ∈ [-1/2, 1/2], that means, θ ∈ [-π/6, π/6] ⇒ 3sin⁻¹x = 3θ and 3θ ∈ [-π/2, π/2]. (1)
3x - x³ = 3sin θ - sin³θ = sin 3θ and 3x - x³ ∈ [-1, 1] since x ∈ [-1/2, 1/2] ⇒ sin⁻¹(3x - x³) = sin⁻¹(sin 3θ) = 3θ. (2)
From (1) and (2), we get, 3sin⁻¹x = sin⁻¹(3x - x³)
Hence Proved.