Trig Identities Question

Prove that:
a) 1 - (sin^6 theta + cos^6 theta) =(3sin^2 theta)(cos^2 theta)?

... LHS

= 1 - ( s⁶ + c⁶ )

= 1 - [ ( s² )³ + ( c² )³ ] ....................... Use : a³ + b³ = (a+b) (a² - ab + b²)

= 1 - [ ( s² + c² ) ( s⁴ - s²c² + c⁴ ) ] .... Here : s² + c² = 1

= 1 - [ ( s² )² + ( c² )² ] + s²c² ............. Use : a² + b² = (a+b)² - 2ab

= 1 - [ ( s² + c² )² - 2s²c² ] + s²c²

= 1 - 1 + 2s²c² + s²c²

= 3s²c²

= RHS .................................. Q.E.D.
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We have:
LHS = 1 - (sin^6θ + cos^6θ)
= (1 - sin^6θ) - cos^6θ, by distributing the negative
= (1 - sin^2θ)(1 + sin^2θ + sin^4θ) - cos^6θ, via difference of cubes
= cos^2θ(1 + sin^2θ + sin^4θ) - cos^6θ, since 1 - sin^2θ = cos^2θ
= cos^2θ(1 + sin^2θ + sin^4θ - cos^4θ), by factoring out cos^2θ
= cos^2θ[1 + sin^2θ + sin^4θ - (1 - sin^2θ)^2], since cos^2θ = 1 - sin^2θ
= cos^2θ(1 + sin^2θ + sin^4θ - sin^4θ + 2sin^2θ - 1)
= 3sin^2θcos^2θ
= RHS.

I hope this helps!