The value of $3{(\sin x-\cos x)}^4+6{(\sin x+\cos x)}^2+4({\sin }^{6}x+{\cos }^{6}x)$ is

 $\begin{array}{l}3{(\sin x-\cos x)}^4+6{(\sin x+\cos x)}^2+4({\sin }^{6}x+{\cos }^{6}x)=3{({\sin }^{2}x+{\cos }^{2}x-2\sin x\cos x)}^2+\\ 6({\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x)+4({\sin }^{2}x+{\cos }^{2}x)({\sin }^{4}x+{\cos }^{4}x-{\sin }^{2}x{\cos }^{2}x)\end{array}$
 $=3{(1-2\sin x\cos x)}^2+6(1+2\sin x\cos x)+4({\sin }^{4}x+{\cos }^{4}x)-4{\sin }^{2}x{\cos }^{2}x$
 $=3(1+4{\sin }^{2}x{\cos }^{2}x-4\sin x\cos x)+6+12\sin x\cos x+4({\sin }^{4}x+{\cos }^{4}x)-4{\sin }^{2}x{\cos }^{2}x$
 $=9+4({\sin }^{4}x+{\cos }^{4}x+2{\sin }^{2}x{\cos }^{2}x)$
 $=9+4{({\sin }^{2}x+{\cos }^{2}x)}^2=9+4=13$