Find the coefficient of `x^5""in(x+3)^8`

Explanation

To find the coefficient of \( x^5 \) in the expansion of \( (x + 3)^8 \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Binomial Expansion Formula**:
The general term in the binomial expansion of \( (p + q)^n \) is given by:
\[ T_r = \binom{n}{r} p^{n-r} q^r \]
where \( \binom{n}{r} \) is the binomial coefficient.

2. **Apply the Formula to Our Problem**:
In our case, \( p = x \), \( q = 3 \), and \( n = 8 \). Thus, the general term \( T_r \) for \( (x + 3)^8 \) is:
\[ T_r = \binom{8}{r} x^{8-r} \cdot 3^r \]

3. **Find the Value of \( r \) for \( x^5 \)**:
We want the term where the power of \( x \) is 5. This means we need to solve for \( r \) in the equation:
\[ 8 - r = 5 \]
Solving this gives:
\[ r = 3 \]

4. **Substitute \( r \) Back into the General Term**:
Now we substitute \( r = 3 \) into the general term:
\[ T_3 = \binom{8}{3} x^{8-3} \cdot 3^3 = \binom{8}{3} x^5 \cdot 27 \]

5. **Calculate the Binomial Coefficient \( \binom{8}{3} \)**:
The binomial coefficient \( \binom{8}{3} \) is calculated as:
\[ \binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8!}{3!5!} \]
Calculating this gives:
\[ \binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56 \]

6. **Combine the Results to Find the Coefficient**:
Now we can find the coefficient of \( x^5 \):
\[ \text{Coefficient of } x^5 = \binom{8}{3} \cdot 3^3 = 56 \cdot 27 \]
Calculating \( 56 \cdot 27 \):
\[ 56 \cdot 27 = 1512 \]

### Final Answer:
The coefficient of \( x^5 \) in the expansion of \( (x + 3)^8 \) is **1512**.