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.