<s>[INST] How to solve this trigonometry equation? Show that $3(\sin θ - \cos θ)^4 + 6(\sin θ + \cos θ)^2 + 4(\sin^6 θ + \cos^6 θ)$ is independent $θ$.
This question was there in one of the cbse  sample papers  for class x . Tried many methods but couldn't solve. 
I got the solution until I found the equation $= 13$ but couldn't understand the independent $θ$ part. It'd be great if you could explain that part. Thanks! 
P.S it's question no.31. 

PPS. A big thank you to everyone who helped
 [/INST] HINT:
$$\sin^6\theta+\cos^6\theta=(\sin^2\theta)^3+(\cos^2\theta)^3$$
$$=(\cos^2\theta+\sin^2\theta)^3-3(\cos^2\theta\sin^2\theta)(\cos^2\theta+\sin^2\theta)=1-3(\cos^2\theta\sin^2\theta)$$
Now,
$$(\sin\theta\pm\cos\theta)^2=\cos^2\theta+\sin^2\theta\pm2\sin\theta\cos\theta=1\pm2\sin\theta\cos\theta$$
$$(\sin\theta-\cos\theta)^4=(1-2\sin\theta\cos\theta)^2=1+4\sin^2\theta\cos^2\theta-4\sin\theta\cos\theta$$
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