To solve this problem, we need to use the Maclaurin series for $\cos(x)$ and apply it to the function $f(x) = x^3 \cos(5x)$.

Step-by-Step Solution

1. Maclaurin Series for $\cos(x)$: The Maclaurin series expansion for $\cos(x)$ is: $\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots$

2. Maclaurin Series for $\cos(5x)$: To find the series for $\cos(5x)$, substitute $5x$ for $x$ in the series: $\cos(5x) = 1 - \frac{(5x)^2}{2!} + \frac{(5x)^4}{4!} - \frac{(5x)^6}{6!} + \dots$ Simplifying each term: $\cos(5x) = 1 - \frac{25x^2}{2} + \frac{625x^4}{24} - \frac{15625x^6}{720} + \dots$

3. Multiply by $x^3$: Now, multiply each term in the series for $\cos(5x)$ by $x^3$ to get the series for $f(x) = x^3 \cos(5x)$: $f(x) = x^3 \left(1 - \frac{25x^2}{2} + \frac{625x^4}{24} - \dots \right)$ Expanding this, we get: $f(x) = x^3 - \frac{25x^5}{2} + \frac{625x^7}{24} - \dots$

4. Identify the First Three Non-Zero Terms: The first three non-zero terms in this series are: $x^3 - \frac{25x^5}{2} + \frac{625x^7}{24}$

Conclusion

The correct answer is: $x^3 - \frac{25x^5}{2} + \frac{625x^7}{24}$ This matches the second option in the provided answer choices.

Would you like a detailed explanation of each term or any other questions? Here are some related questions for practice:

1. What is the general formula for the Maclaurin series expansion?
2. How would you find the Maclaurin series for $\sin(x)$ and use it in a similar problem?
3. How does substituting $5x$ in $\cos(x)$ change the Maclaurin series?
4. Can we approximate $f(x)$ using just the first two non-zero terms?
5. What are the limitations of using Maclaurin series expansions?

Tip: Remember, multiplying the series term-by-term is an efficient way to adjust a known series expansion for functions like $x^3 \cos(5x)$.