<s>[INST] If $\cos^4 \theta −\sin^4 \theta = x$. Find $\cos^6 \theta − \sin^6 \theta $ in terms of $x$. Given $\cos^4 \theta −\sin^4  \theta = x$ , I've to find the value of $\cos^6 \theta − \sin^6 \theta $ .
Here is what I did:
$\cos^4 \theta −\sin^4  \theta = x$.
($\cos^2 \theta −\sin^2  \theta)(\cos^2 \theta +\sin^2  \theta) = x$
Thus 
($\cos^2 \theta −\sin^2  \theta)=x$ ,
so $\cos 2\theta=x$ .
Now $x^3=(\cos^2 \theta −\sin^2  \theta)^3=\cos^6 \theta-\sin^6 \theta +3 \sin^4 \theta \cos^2 \theta -3 \sin^2 \theta \cos^4 \theta $
So if I can find the value of $3 \sin^4 \theta \cos^2 \theta -3 \sin^2 \theta \cos^4 \theta $ in terms of $x$ , the question is solved. But how to do that ?
 [/INST] It would be easier if you decompose $\cos^6 \theta -\sin^6 \theta $ into $(\cos^2 \theta -\sin^2 \theta)(\cos^4 \theta +\cos^2 \theta \sin^2 \theta+\sin^4 \theta)$
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