Prove that \[ 2(\sin^6x+\cos^6x) – 3(\sin^4x+\cos^4x) +1 =0 \]

Proof: Using \[ a^3+b^3=(a+b)^3-3ab(a+b) \] with \[ a=\sin^2x,\quad b=\cos^2x \] we get \[ \sin^6x+\cos^6x = (\sin^2x+\cos^2x)^3 – 3\sin^2x\cos^2x(\sin^2x+\cos^2x) \] Using \[ \sin^2x+\cos^2x=1 \] therefore, \[ \sin^6x+\cos^6x = 1-3\sin^2x\cos^2x \] Also, \[ \sin^4x+\cos^4x = (\sin^2x+\cos^2x)^2 – 2\sin^2x\cos^2x \] \[ =1-2\sin^2x\cos^2x \] Substituting these values into LHS: \[ LHS = 2(1-3\sin^2x\cos^2x) – 3(1-2\sin^2x\cos^2x) +1 \] Expanding: \[ = 2-6\sin^2x\cos^2x -3+6\sin^2x\cos^2x +1 \] Combining like terms: \[ =(2-3+1) + (-6\sin^2x\cos^2x+6\sin^2x\cos^2x) \] \[ =0+0 \] \[ =0 \] Hence proved, \[ \boxed{ 2(\sin^6x+\cos^6x) – 3(\sin^4x+\cos^4x) +1 =0 } \]