Question

 Prove that 2(sin^6 theta + cos^6 theta) - 3(sin^4 
theta+ cos^4 theta) + 1 = 0

Answer 1

let theta be x
LHS =2((sin^2x)^3 + (cos^2x)^3) - 3sin^4x - 3cos^4x + 1
   = 2[ sin^2x+cos^2x][sin^4x- sin^2x * cos^2x +cos^4x]-3sin^4x-3cos^4x+1
apply a^3+b^3=(a+b)(a^2-ab+b^2)
also sin^2x+cos^2x=1
s0
=2sin^4x-2sin^2xcos^2x+2cos^4x-3sin^4x-3cos^4x+1
=[-sin^4x-cos^4x-2sin^2xcos^2x]+1
[sin^2x+cos^2x]^2=sin^4x+2sin^2xcos^2x+cos^4x
so
=-[sin^2x+cos^2x]^2+1
=-1+1
=0
LHS=RHS