([b]4[/b]) Let $ f(x) \equal{} \sin^6\left(\frac {x}{4}\right) \plus{} \cos^6\left(\frac {x}{4}\right)$ for all real numbers $ x$. Determine $ f^{(2008)}(0)$ (i.e., $ f$ differentiated $ 2008$ times and then evaluated at $ x \equal{} 0$).
1. We start with the function \( f(x) = \sin^6\left(\frac{x}{4}\right) + \cos^6\left(\frac{x}{4}\right) \).

2. To simplify \( \sin^6\left(\frac{x}{4}\right) + \cos^6\left(\frac{x}{4}\right) \), we use the identity:
   \[
   \sin^6 y + \cos^6 y = (\sin^2 y + \cos^2 y)(\sin^4 y - \sin^2 y \cos^2 y + \cos^4 y)
   \]
   Since \( \sin^2 y + \cos^2 y = 1 \), we have:
   \[
   \sin^6 y + \cos^6 y = \sin^4 y - \sin^2 y \cos^2 y + \cos^4 y
   \]

3. Next, we use the identity for \( \sin^4 y + \cos^4 y \):
   \[
   \sin^4 y + \cos^4 y = (\sin^2 y + \cos^2 y)^2 - 2\sin^2 y \cos^2 y = 1 - 2\sin^2 y \cos^2 y
   \]
   Therefore:
   \[
   \sin^6 y + \cos^6 y = 1 - 3\sin^2 y \cos^2 y
   \]

4. Using the double-angle identity \( \sin^2 y \cos^2 y = \frac{1}{4} \sin^2(2y) \), we get:
   \[
   \sin^6 y + \cos^6 y = 1 - \frac{3}{4} \sin^2(2y)
   \]

5. Applying this to our function \( f(x) \) with \( y = \frac{x}{4} \), we have:
   \[
   f(x) = 1 - \frac{3}{4} \sin^2\left(\frac{x}{2}\right)
   \]

6. Using the identity \( \sin^2 z = \frac{1 - \cos(2z)}{2} \), we get:
   \[
   \sin^2\left(\frac{x}{2}\right) = \frac{1 - \cos(x)}{2}
   \]
   Therefore:
   \[
   f(x) = 1 - \frac{3}{4} \cdot \frac{1 - \cos(x)}{2} = 1 - \frac{3}{8} (1 - \cos(x)) = \frac{5}{8} + \frac{3}{8} \cos(x)
   \]

7. Now, we need to find the 2008th derivative of \( f(x) \) at \( x = 0 \). The function \( f(x) = \frac{5}{8} + \frac{3}{8} \cos(x) \) is a linear combination of constants and \( \cos(x) \).

8. The 2008th derivative of a constant is zero. The 2008th derivative of \( \cos(x) \) is:
   \[
   \frac{d^{2008}}{dx^{2008}} \cos(x) = \cos(x + \frac{2008\pi}{2}) = \cos(x + 1004\pi) = \cos(x)
   \]
   Since \( \cos(x) \) is periodic with period \( 2\pi \), the 2008th derivative of \( \cos(x) \) is \( \cos(x) \).

9. Evaluating at \( x = 0 \):
   \[
   f^{(2008)}(0) = \frac{3}{8} \cos(0) = \frac{3}{8} \cdot 1 = \frac{3}{8}
   \]

The final answer is \(\boxed{\frac{3}{8}}\)