Express cos6θ in terms of powers of cosθ using De Moivre's Theorem.

De Moivre's Theorem states that $(\cos \theta + i \sin \theta)^n = \cos(n\theta) + i \sin(n\theta)$.
For $n=6$, we have:

$$(\cos \theta + i \sin \theta)^6 = \cos(6\theta) + i \sin(6\theta)$$

Let $c = \cos \theta$ and $s = \sin \theta$.

$$(c + is)^6 = \binom{6}{0}c^6 + \binom{6}{1}c^5(is) + \binom{6}{2}c^4(is)^2 + \binom{6}{3}c^3(is)^3 + \binom{6}{4}c^2(is)^4 + \binom{6}{5}c(is)^5 + \binom{6}{6}(is)^6$$

Substitute the binomial coefficients $\binom{6}{0}=1, \binom{6}{1}=6, \binom{6}{2}=15, \binom{6}{3}=20, \binom{6}{4}=15, \binom{6}{5}=6, \binom{6}{6}=1$ and powers of $i$ ($i^2=-1, i^3=-i, i^4=1, i^5=i, i^6=-1$):

$$= 1 \cdot c^6 + 6 \cdot c^5 (is) + 15 \cdot c^4 (-s^2) + 20 \cdot c^3 (-is^3) + 15 \cdot c^2 (s^4) + 6 \cdot c (is^5) + 1 \cdot (-s^6)$$ $$= c^6 + 6ic^5s - 15c^4s^2 - 20ic^3s^3 + 15c^2s^4 + 6ics^5 - s^6$$

The real terms are those without $i$:

$$\cos(6\theta) = c^6 - 15c^4s^2 + 15c^2s^4 - s^6$$

Substitute back $c = \cos \theta$ and $s = \sin \theta$:

$$\cos(6\theta) = \cos^6 \theta - 15 \cos^4 \theta \sin^2 \theta + 15 \cos^2 \theta \sin^4 \theta - \sin^6 \theta$$

Let $x = \cos^2 \theta$.

$$\cos(6\theta) = x^3 - 15x^2(1-x) + 15x(1-x)^2 - (1-x)^3$$ $$= x^3 - 15x^2 + 15x^3 + 15x(1 - 2x + x^2) - (1 - 3x + 3x^2 - x^3)$$ $$= x^3 - 15x^2 + 15x^3 + 15x - 30x^2 + 15x^3 - 1 + 3x - 3x^2 + x^3$$

Group terms by powers of $x$:

$$x^3: (1 + 15 + 15 + 1)x^3 = 32x^3$$ $$x^2: (-15 - 30 - 3)x^2 = -48x^2$$ $$x: (15 + 3)x = 18x$$ $$Constant: -1$$

Substitute $x = \cos^2 \theta$ back into the simplified expression:

$$\cos(6\theta) = 32 \cos^6 \theta - 48 \cos^4 \theta + 18 \cos^2 \theta - 1$$

The final answer is $32 \cos^6 \theta - 48 \cos^4 \theta + 18 \cos^2 \theta - 1$.