To find the Taylor series of $f(x) = x^{1/3}$ about $a = 8$, we use the formula for the Taylor series expansion:

$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n$

Step 1: Compute the derivatives of $f(x) = x^{1/3}$.

1. First derivative $f'(x)$: $f'(x) = \frac{1}{3}x^{-2/3}$

2. Second derivative $f''(x)$: $f''(x) = -\frac{2}{9}x^{-5/3}$

3. Third derivative $f^{(3)}(x)$: $f^{(3)}(x) = \frac{10}{27}x^{-8/3}$

Step 2: Evaluate the function and derivatives at $x = 8$.

1. Evaluate $f(8)$: $f(8) = 8^{1/3} = 2$

2. Evaluate $f'(8)$: $f'(8) = \frac{1}{3} \cdot 8^{-2/3} = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}$

3. Evaluate $f''(8)$: $f''(8) = -\frac{2}{9} \cdot 8^{-5/3} = -\frac{2}{9} \cdot \frac{1}{32} = -\frac{1}{144}$

4. Evaluate $f^{(3)}(8)$: $f^{(3)}(8) = \frac{10}{27} \cdot 8^{-8/3} = \frac{10}{27} \cdot \frac{1}{256} = \frac{10}{6912}$

Step 3: Write out the Taylor series up to $(x - 8)^2$.

The Taylor series expansion is:

$f(x) \approx f(8) + f'(8)(x - 8) + \frac{f''(8)}{2!}(x - 8)^2$

Substitute the values we calculated:

$f(x) \approx 2 + \frac{1}{12}(x - 8) - \frac{1}{144} \cdot \frac{(x - 8)^2}{2}$

Simplifying:

$f(x) \approx 2 + \frac{1}{12}(x - 8) - \frac{1}{288}(x - 8)^2$

This is the Taylor series of $f(x) = x^{1/3}$ about $x = 8$ up to the $(x - 8)^2$ term.

Would you like any further clarification or more detailed steps?

Follow-up Questions:

1. What would the next term in the series be if we included the third derivative?
2. How would the series change if we expanded around a different value of $a$?
3. How does the accuracy of the approximation depend on the number of terms in the Taylor series?
4. Could this series be used to approximate cube roots for values near $x = 8$?
5. What is the geometric interpretation of the Taylor series approximation?

Tip:

When expanding functions using Taylor series, it's helpful to calculate a few higher-order derivatives, even if they're not immediately used, as they can show how quickly the series converges.