Prove that : (1+cotθ+tanθ)(sinθ-cosθ)/sec³θ-cosec³θ . . .

Question : Prove that : 

[(1+cotθ+tanθ)(sinθ-cosθ)]/[sec³θ-cosec³θ] 

= sin²θcos²θ

Doubt by Saksham

Solution : 

[∵ (a³-b³)=(a-b)(a²+b²+ab)]
= sin³θcos³θ / sinθcosθ
= sin²θcos²θ

= RHS

LHS = RHS 
Hence Proved.