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## Answers

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