1. The problem asks to produce the power series for $\cos^2(2x)$ up to the term in $x^6$. 2. Recall the double-angle identity: $$\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}$$ Apply it with $\theta = 2x$: $$\cos^2(2x) = \frac{1 + \cos(4x)}{2}$$ 3. We need the power series for $\cos(4x)$ up to $x^6$. The Maclaurin series for $\cos z$ is: $$\cos z = \sum_{n=0}^{\infty} (-1)^n \frac{z^{2n}}{(2n)!} = 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \frac{z^6}{6!} + \cdots$$ 4. Substitute $z=4x$: $$\cos(4x) = 1 - \frac{(4x)^2}{2!} + \frac{(4x)^4}{4!} - \frac{(4x)^6}{6!} + \cdots$$ Calculate terms explicitly: $$= 1 - \frac{16x^2}{2} + \frac{256x^4}{24} - \frac{4096x^6}{720} + \cdots$$ $$= 1 - 8x^2 + \frac{256}{24} x^4 - \frac{4096}{720} x^6 + \cdots$$ Simplify fractions: $$= 1 - 8x^2 + \frac{32}{3} x^4 - \frac{512}{90} x^6 + \cdots$$ $$= 1 - 8x^2 + \frac{32}{3} x^4 - \frac{512}{90} x^6 + \cdots$$ 5. Now plug this back into the formula for $\cos^2(2x)$: $$\cos^2(2x) = \frac{1 + \cos(4x)}{2} = \frac{1 + \left(1 - 8x^2 + \frac{32}{3} x^4 - \frac{512}{90} x^6\right)}{2}$$ $$= \frac{2 - 8x^2 + \frac{32}{3} x^4 - \frac{512}{90} x^6}{2}$$ 6. Divide each term by 2: $$= 1 - 4x^2 + \frac{16}{3} x^4 - \frac{256}{90} x^6$$ Simplify the last fraction: $$\frac{256}{90} = \frac{128}{45}$$ 7. Final power series for $\cos^2(2x)$ up to $x^6$ is: $$\boxed{1 - 4x^2 + \frac{16}{3} x^4 - \frac{128}{45} x^6}$$ This completes the required power series expansion.