Solve $3(\sin x-\cos x)^{4}+6(\sin x+\cos x)^{2}+4(\sin^{6} x+\cos^{6} x)$ Solve $3(\sin x-\cos x)^{4}+6(\sin x+\cos x)^{2}+4(\sin^{6} x+\cos^{6} x)$
$a.)\ 1 \\
\color{green}{b.)\ 13} \\
c.)\ 15 \\
d.)\ 16 $
$ 3(\sin x-\cos x)^{4}+6(\sin x+\cos x)^{2}+4(\sin^{6} x+\cos^{6} x)\\
=3(1-\sin (2x))^{2}+6(1+\sin 2x)+4(\sin^{6} x+\cos^{6} x)\\
=3(1-2\sin (2x)+\sin^{2} (2x))+6(1+\sin 2x)+4(\sin^{6} x+\cos^{6} x)\\
=9+3\sin^{2} 2x+4(\sin^{6} x+\cos^{6} x)\\
=9+12\sin^{2}x\times \cos^{2}x+4(\sin^{6} x+\cos^{6} x)\\$
I am stucked.
I look for a short and simple way.
I have studied maths upto $12$th grade.
 A: The answer should be the same for all $x$. So plug in $x = 0$, 
$$3(-1)^4+6(1)^2+4(1)^6 = 13$$
A: Note that
$$\begin{align}&\sin^6x+\cos^6x\\&=(\sin^2x+\cos^2x)(\sin^4x-\sin^2x\cos^2x+\cos^4x)\\&=1\cdot ((\sin^2x+\cos^2x)^2-3\sin^2x\cos^2x)\\&=1-3\sin^2x\cos^2x\end{align}$$
A: $$3\left(\sin(x)-\cos(x)\right)^4+6\left(\sin(x)+\cos(x)\right)^2+4\left(\sin^6(x)+\cos^6(x)\right)=$$
$$3\left(-\sqrt{2}\sin\left(\frac{\pi}{4}-x\right)\right)^4+6\left(\sqrt{2}\sin\left(\frac{\pi}{4}+x\right)\right)^2+4\left(\frac{3\cos(4x)+5}{8}\right)=$$
$$3\left(-\sqrt{2}\sin\left(\frac{\pi}{4}-x\right)\right)^4+6\left(\sqrt{2}\sin\left(\frac{\pi}{4}+x\right)\right)^2+\frac{12\cos(4x)+20}{8}=$$
$$3\left(-\sqrt{2}\sin\left(\frac{\pi}{4}-x\right)\right)^4+6\left(\sqrt{2}\sin\left(\frac{\pi}{4}+x\right)\right)^2+\frac{3\cos(4x)+5}{2}=$$
$$3\left(4\sin^4\left(\frac{\pi}{4}-x\right)\right)+6\left(2\sin^2\left(\frac{\pi}{4}+x\right)\right)+\frac{3\cos(4x)+5}{2}=$$
$$12\sin^4\left(\frac{\pi}{4}-x\right)+12\sin^2\left(\frac{\pi}{4}+x\right)+\frac{3\cos(4x)+5}{2}=$$
$$\frac{12\sin(2x)+3\cos(4x)+17}{2}-\frac{12\sin(2x)+3\cos(4x)-9}{2}=$$
$$\frac{12\sin(2x)+3\cos(4x)+17-12\sin(2x)-3\cos(4x)+9}{2}=$$
$$\frac{0+3\cos(4x)+17-0-3\cos(4x)+9}{2}=$$
$$\frac{3\cos(4x)+17-3\cos(4x)+9}{2}=$$
$$\frac{0+17-0+9}{2}=$$
$$\frac{17+9}{2}=$$
$$\frac{26}{2}=13$$
A: 
$get  x=0 \Rightarrow 3+6+4=13$

