Prove that \[ 3(\sin x-\cos x)^4 + 6(\sin x+\cos x)^2 + 4(\sin^6x+\cos^6x) =13 \]

Proof: Using \[ (\sin x-\cos x)^2 = \sin^2x+\cos^2x-2\sin x\cos x \] and \[ \sin^2x+\cos^2x=1 \] we get \[ (\sin x-\cos x)^2 = 1-2\sin x\cos x \] Therefore, \[ (\sin x-\cos x)^4 = (1-2\sin x\cos x)^2 \] \[ = 1-4\sin x\cos x+4\sin^2x\cos^2x \] Also, \[ (\sin x+\cos x)^2 = 1+2\sin x\cos x \] Now, \[ \sin^6x+\cos^6x = (\sin^2x)^3+(\cos^2x)^3 \] Using \[ a^3+b^3=(a+b)^3-3ab(a+b) \] 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, \[ \sin^6x+\cos^6x = 1-3\sin^2x\cos^2x \] Substituting all values in LHS: \[ LHS = 3(1-4\sin x\cos x+4\sin^2x\cos^2x) \] \[ +6(1+2\sin x\cos x) \] \[ +4(1-3\sin^2x\cos^2x) \] Expanding: \[ = 3-12\sin x\cos x+12\sin^2x\cos^2x \] \[ +6+12\sin x\cos x \] \[ +4-12\sin^2x\cos^2x \] Combining like terms: \[ =3+6+4 \] \[ =13 \] Hence proved, \[ \boxed{ 3(\sin x-\cos x)^4 + 6(\sin x+\cos x)^2 + 4(\sin^6x+\cos^6x) =13 } \]