<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] For typing ease, let $a=\cos^2\theta$ and $b=\sin^2\theta$. Thus $a+b=1$. We are told that $a^2-b^2=x$, or equivalently that $a-b=x$. 
We want $a^3-b^3$.  It will be enough to find $a^2+ab+b^2$. Note the identity
$$4(a^2+ab+b^2)=3(a+b)^2+(a-b)^2.$$
Remark: For other problems of a similar character, it might be preferable to extract $ab$ from the identity $4ab=(a+b)^2-(a-b)^2$, and use standard techniques for expressing symmetric functions of $a$ and $b$ in terms of the elementary symmetric functions $a+b$ and $ab$.
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