Question

\[ 3(\sin x-\cos x)^4 +6(\sin x+\cos x)^2 +4(\sin^6x+\cos^6x) \]

Solution

Using identities

\[ (\sin x-\cos x)^2 = 1-2\sin x\cos x \]

\[ (\sin x+\cos x)^2 = 1+2\sin x\cos x \]

Let

\[ \sin x\cos x=t \]

Then

\[ 3(1-2t)^2+6(1+2t) +4(\sin^6x+\cos^6x) \]

Now,

\[ \sin^6x+\cos^6x = (\sin^2x+\cos^2x)^3 -3\sin^2x\cos^2x(\sin^2x+\cos^2x) \]

\[ =1-3t^2 \]

Substituting,

\[ 3(1-4t+4t^2)+6+12t+4(1-3t^2) \]

\[ =3-12t+12t^2+6+12t+4-12t^2 \]

\[ =13 \]

Answer

\[ \boxed{13} \]