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 follow these steps:

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term in the binomial 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 \( n = 10 \).

2. **Write the General Term**:
Substituting the values into the formula, we get:
\[ T_{r+1} = \binom{10}{r} (2x^3)^{10-r} \left(-\frac{1}{3x^3}\right)^r \]

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

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

5. **Substitute \( r \) into the General Term**:
Now, substitute \( r = 8 \) back into the general term:
\[ T_{9} = \binom{10}{8} (-1)^8 \frac{2^{10-8}}{3^8} x^{6} \]
Simplifying this gives:
\[ T_{9} = \binom{10}{8} \frac{2^2}{3^8} x^6 \]

6. **Calculate the Coefficient**:
The binomial coefficient \( \binom{10}{8} = \binom{10}{2} = 45 \). Therefore, the coefficient of \( x^6 \) is:
\[ 45 \cdot \frac{4}{3^8} = \frac{180}{6561} \]

7. **Final Result**:
Thus, the coefficient of \( x^6 \) in the expansion is:
\[ \frac{180}{6561} \]