Trig identity: sin2θ+cos2θ=1 is used below.
a3+b3=(a+b)(a2-ab+b2), so let a=sin2θ, b=cos2θ, then:
sin6θ+cos6θ=(sin2θ+cos2θ)(sin4θ-sin2θcos2θ+cos4θ)=sin4θ-sin2θcos2θ+cos4θ.
2(sin6θ+cos6θ)=2(sin4θ-sin2θcos2θ+cos4θ);
subtract 3(sin4θ+cos4θ)=-sin4θ-cos4θ-2sin2θcos2θ=-(sin2θ+cos2θ)2=-1;
add 1=0 QED