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The given function is \( f(x) = x^3 \), and we are tasked with finding the Taylor series expansion around \( x = 1 \).

### Step 1: Taylor Series Definition

The Taylor series of a function \( f(x) \) about \( x = a \) is given by:
\[
f(x) = \sum_{n=0}^{\infty} c_n (x - a)^n
\]
where \( c_n \) represents the coefficients:
\[
c_n = \frac{f^{(n)}(a)}{n!}
\]
Here, \( a = 1 \) and \( f(x) = x^3 \).

### Step 2: Calculate Derivatives

1. \( f(x) = x^3 \)
2. First derivative: \( f'(x) = 3x^2 \)
3. Second derivative: \( f''(x) = 6x \)
4. Third derivative: \( f^{(3)}(x) = 6 \)
5. Fourth derivative and higher: \( f^{(n)}(x) = 0 \) for \( n \geq 4 \)

### Step 3: Evaluate Derivatives at \( x = 1 \)

We evaluate each derivative at \( x = 1 \):

1. \( f(1) = 1^3 = 1 \)
2. \( f'(1) = 3(1)^2 = 3 \)
3. \( f''(1) = 6(1) = 6 \)
4. \( f^{(3)}(1) = 6 \)
5. \( f^{(n)}(1) = 0 \) for \( n \geq 4 \)

### Step 4: Find Coefficients

Using the formula \( c_n = \frac{f^{(n)}(1)}{n!} \), we compute the coefficients:

1. \( c_0 = \frac{f(1)}{0!} = \frac{1}{1} = 1 \)
2. \( c_1 = \frac{f'(1)}{1!} = \frac{3}{1} = 3 \)
3. \( c_2 = \frac{f''(1)}{2!} = \frac{6}{2} = 3 \)
4. \( c_3 = \frac{f^{(3)}(1)}{3!} = \frac{6}{6} = 1 \)
5. \( c_4 = \frac{f^{(4)}(1)}{4!} = \frac{0}{24} = 0 \)

### Conclusion

The first few coefficients are:
- \( c_0 = 1 \)
- \( c_1 = 3 \)
- \( c_2 = 3 \)
- \( c_3 = 1 \)
- \( c_4 = 0 \)

Would you like more details on how the Taylor series expansion is derived?

### Related Questions
1. How do we compute higher-order coefficients in Taylor series?
2. Can the Taylor series be used to approximate \( f(x) = x^3 \) near other points besides \( x = 1 \)?
3. How does the radius of convergence affect the accuracy of a Taylor series?
4. What happens if we expand \( f(x) = x^3 \) around a different point, say \( x = 0 \)?
5. What is the geometric interpretation of the Taylor series for a cubic function?

### Tip:
To approximate functions using Taylor series effectively, focus on finding the terms up to the degree where further terms don't significantly contribute to the result within your desired range of accuracy.

The Taylor series for f(x) = x^3 at 1 is ∑ c_n (x - 1)^n. Find the first few coefficients.

Learn how to compute the Taylor series expansion of the function f(x) = x^3 around the point x = 1. This detailed solution covers key calculus concepts, including the calculation of derivatives and coefficients using the Taylor series formula. Suitable for students studying undergraduate-level calculus.

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