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To solve this problem, we need to find the coefficient of \( x^3 \) in the expansion of \( (5 + 3x)^5 \).

We can use the **Binomial Theorem** for expanding expressions of the form \( (a + b)^n \). The Binomial Theorem states:

\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\]

In this case, \( a = 5 \), \( b = 3x \), and \( n = 5 \).

To find the term involving \( x^3 \), we need the binomial term where the power of \( x \) is 3. This corresponds to the term where \( k = 3 \), since the power of \( x \) is determined by \( k \).

The general term in the binomial expansion is given by:

\[
T_k = \binom{n}{k} a^{n-k} b^k
\]

For \( k = 3 \):

\[
T_3 = \binom{5}{3} 5^{5-3} (3x)^3
\]

Simplifying this:

\[
T_3 = \binom{5}{3} 5^2 (3x)^3
\]
\[
T_3 = \binom{5}{3} 25 \cdot 27x^3
\]
\[
T_3 = \binom{5}{3} 25 \cdot 27 x^3
\]

Now, calculate \( \binom{5}{3} \):

\[
\binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10
\]

So:

\[
T_3 = 10 \cdot 25 \cdot 27 x^3 = 6750 x^3
\]

Thus, the coefficient of \( x^3 \) is **6750**.

The correct answer is **d. 6750**.

Do you have any questions or would you like more details?

### Follow-up questions:
1. How is the binomial coefficient calculated in the binomial expansion?
2. What would be the coefficient of \( x^4 \) in the same expansion?
3. How would the expansion change if the exponent was higher, like \( (5 + 3x)^6 \)?
4. Can the binomial theorem be used for negative exponents?
5. How is the binomial theorem useful in probability theory?

### Tip:
The binomial theorem is extremely useful for expanding polynomials and calculating combinations without fully expanding each term manually.

Find the coefficient of x^3 in the expansion of (5 + 3x)^5.

Learn how to find the coefficient of x^3 in the expansion of (5 + 3x)^5 using the Binomial Theorem. This problem covers key concepts in binomial expansion and algebra, and is suitable for students in Grades 9-12.

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