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 ?
 A: $$
\cos^6\theta-\sin^6\theta = \left ( \cos^2 \theta\right )^3 - \left (\sin^2 \theta \right )^3 = \\
= \left( \cos^2 \theta - \sin^2 \theta\right ) \left(\cos^4 \theta + \sin^2\theta \cos^2\theta + \sin^4\theta \right ) = \\
= x \left ( \cos^4 \theta - 2\cos^2\theta\sin^2\theta + \sin^4 \theta + 3 \cos^2\theta\sin^2\theta\right ) = \\
= x \left( \left(\cos^2\theta - \sin^2 \theta \right )^2 + \frac 34 \sin^22\theta\right ) = x \left( x^2 + \frac 34 \left( 1-\cos^2 2\theta\right )\right ) = \\
= x \left(x^2 + \frac 34 (1-x^2) \right ) = \frac x4 \left(x^2+3 \right )
$$
A: 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 $ab$.
A: It would be easier if you decompose $\cos^6 \theta -\sin^6 \theta $ into $(\cos^2 \theta -\sin^2 \theta)(\cos^4 \theta +\cos^2 \theta \sin^2 \theta+\sin^4 \theta)$
A: As already found $\cos^2\theta-\sin^2\theta=x,\implies \cos2\theta=x$
Using $a^3-b^3=(a-b)^3+3ab(a-b),$
$$\cos^6\theta-\sin^6\theta=(\cos^2\theta)^3-(\sin^2\theta)^3$$
$$=(\cos^2\theta-\sin^2\theta)^3+3\cos^2\theta\sin^2\theta(\cos^2\theta-\sin^2\theta) $$
$$=x^3+3x\cos^2\theta\sin^2\theta$$
$$=x^3+\frac{3x}4\sin^22\theta\text{ as }\sin2A=2\sin A\cos A$$
$$=x^3+\frac{3x}4(1-\cos^22\theta)$$
$$=x^3+\frac{3x}4(1-x^2)=\frac{4x^3+3x-3x^3}4=\frac{x(x^2+3)}4$$
A: You have
$$\cos^2 \theta + \sin^2 \theta = 1$$
$$\cos^2 \theta - \sin^2 \theta = x$$
Adding and subtracting the two equations gives
$$\cos^2 \theta = {1 + x \over 2}$$
$$\sin^2 \theta = {1 - x \over 2}$$
Substituting you have
$$\cos^6 \theta - \sin^6 \theta = \bigg({1 + x \over 2}\bigg)^3 - \bigg({1 - x \over 2}\bigg)^3$$
$$ =  {3 \over 4} x + {1 \over 4} x^3$$
