Show that $\frac{d}{dx}(\tan^{3} 2x) = 6\sec^{4} 2x - 6\sec^{2} 2x$. Hence, find the exact value of $\displaystyle \int_{0}^{\pi/8} \sec^{4} 2x\, dx$.

1. Differentiate the expression using the chain rule

$\frac{d}{dx}(\tan^{3} 2x) = 3\tan^{2}(2x) \cdot \frac{d}{dx}(\tan 2x)$
$= 3\tan^{2}(2x) \cdot 2\sec^{2}(2x)$
$= 6\tan^{2}(2x)\sec^{2}(2x)$

2. Use a trigonometric identity to express the derivative in terms of secant only

Using the identity $\tan^{2} \theta = \sec^{2} \theta - 1$:
$= 6(\sec^{2} 2x - 1)\sec^{2} 2x$
$= 6\sec^{4} 2x - 6\sec^{2} 2x$ (Shown)

1. Use the result from part (a) to set up an integral equation.

From part (a), we have $\frac{d}{dx}(\tan^{3} 2x) = 6\sec^{4} 2x - 6\sec^{2} 2x$.
Integrating both sides with respect to $x$ gives:
$\int (6\sec^{4} 2x - 6\sec^{2} 2x)\, dx = \tan^{3} 2x$

2. Rearrange the equation to make the integral of $\sec^4 2x$ the subject.

$6\int \sec^{4} 2x\, dx - 6\int \sec^{2} 2x\, dx = \tan^{3} 2x$
$6\int \sec^{4} 2x\, dx = \tan^{3} 2x + 6\int \sec^{2} 2x\, dx$
$\int \sec^{4} 2x\, dx = \frac{1}{6}\tan^{3} 2x + \int \sec^{2} 2x\, dx$

3. Integrate $\sec^2 2x$.

Since $\int \sec^{2} 2x\, dx = \frac{1}{2}\tan 2x$, we have:
$\int \sec^{4} 2x\, dx = \frac{1}{6}\tan^{3} 2x + \frac{1}{2}\tan 2x$

4. Apply the limits of integration.

$\int_{0}^{\pi/8} \sec^{4} 2x\, dx = \left[ \frac{1}{6}\tan^{3} 2x + \frac{1}{2}\tan 2x \right]_{0}^{\pi/8}$

5. Evaluate the exact value.

$= \left( \frac{1}{6}\tan^{3}\left(\frac{\pi}{4}\right) + \frac{1}{2}\tan\left(\frac{\pi}{4}\right) \right) - \left( \frac{1}{6}\tan^{3}(0) + \frac{1}{2}\tan(0) \right)$
$= \left( \frac{1}{6}(1)^{3} + \frac{1}{2}(1) \right) - (0)$
$= \frac{1}{6} + \frac{1}{2} = \frac{4}{6} = \frac{2}{3}$