Prove the following with the help of identities:

sin⁶θ + cos⁶θ = 1 – 3sin³θ cos²θ

Using the identity: a³ + b³ = (a + b)(a² + b² – ab)

Here, a = sin² θ and b = cos² θ

⇒ (sin² θ + cos² θ)(sin⁴ θ + cos⁴ θ – sin² θ.cos² θ)

Now, Using: sin²θ + cos²θ = 1 and a² + b² = (a + b)² – 2.a.b

⇒ (1)((sin² θ + cos² θ)² – 2.sin² θ.cos² θ – sin²θ.cos²θ )

⇒ ((1)² – 2.sin² θ.cos² θ – sin² θ.cos² θ)

⇒ (1 – 3.sin² θ.cos² θ)

= R.H.S

Hence, proved.