Find the coefficient of `x^(6)` in the expansion of `(2x^(3)-(1)/(3x^(3)))^(10)`

Explanation

To find the coefficient of \( x^6 \) in the expansion of \( (2x^3 - \frac{1}{3x^3})^{10} \), we can use the Binomial Theorem.

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term in the expansion of \( (a + b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
Here, \( a = 2x^3 \), \( b = -\frac{1}{3x^3} \), and \( n = 10 \). Thus, the general term can be expressed as:
\[ T_{r+1} = \binom{10}{r} (2x^3)^{10-r} \left(-\frac{1}{3x^3}\right)^r \]

2. **Simplify the General Term**:
Expanding \( T_{r+1} \):
\[ T_{r+1} = \binom{10}{r} (2^{10-r} x^{3(10-r)}) \left(-\frac{1}{3^r x^{3r}}\right) \]
This simplifies to:
\[ T_{r+1} = \binom{10}{r} 2^{10-r} (-1)^r \frac{1}{3^r} x^{30 - 3r} \]

3. **Find the Power of \( x \)**:
We want the power of \( x \) to be 6:
\[ 30 - 3r = 6 \]
Solving for \( r \):
\[ 30 - 6 = 3r \implies 24 = 3r \implies r = 8 \]

4. **Substitute \( r \) into the General Term**:
Now we substitute \( r = 8 \) back into the general term to find the coefficient:
\[ T_{9} = \binom{10}{8} 2^{10-8} (-1)^8 \frac{1}{3^8} x^{30 - 3 \cdot 8} \]
This simplifies to:
\[ T_{9} = \binom{10}{8} 2^2 \cdot 1 \cdot \frac{1}{3^8} x^{6} \]

5. **Calculate the Coefficient**:
Now we calculate the coefficient:
\[ \binom{10}{8} = \binom{10}{2} = \frac{10 \times 9}{2 \times 1} = 45 \]
Thus, the coefficient is:
\[ 45 \cdot 4 \cdot \frac{1}{3^8} = \frac{180}{6561} \]

### Final Answer:
The coefficient of \( x^6 \) in the expansion is:
\[ \frac{180}{6561} \]