The term independent of x in `(2x - (1)/(3x) )^6` is

Options

A. `160/9`
B. `80/9`
C. `160/27`
D. `80/3`

Answer

To find the term independent of \( x \) in the expression \( (2x - \frac{1}{3x})^6 \), we can follow these steps:

### Step 1: Identify the General Term
The general term \( T_{r+1} \) in the binomial expansion of \( (a + b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
For our expression, we have \( a = 2x \), \( b = -\frac{1}{3x} \), and \( n = 6 \).

### Step 2: Write the General Term for Our Expression
Substituting the values into the formula, we get:
\[ T_{r+1} = \binom{6}{r} (2x)^{6-r} \left(-\frac{1}{3x}\right)^r \]

### Step 3: Simplify the General Term
Now, simplifying \( T_{r+1} \):
\[ T_{r+1} = \binom{6}{r} (2^{6-r} x^{6-r}) \left(-\frac{1}{3^r x^r}\right) \]
\[ = \binom{6}{r} (-1)^r \frac{2^{6-r}}{3^r} x^{6-r-r} \]
\[ = \binom{6}{r} (-1)^r \frac{2^{6-r}}{3^r} x^{6-2r} \]

### Step 4: Find the Term Independent of \( x \)
To find the term independent of \( x \), we set the exponent of \( x \) to zero:
\[ 6 - 2r = 0 \]
Solving for \( r \):
\[ 2r = 6 \implies r = 3 \]

### Step 5: Substitute \( r \) Back into the General Term
Now, substitute \( r = 3 \) back into the general term:
\[ T_{4} = \binom{6}{3} (-1)^3 \frac{2^{6-3}}{3^3} x^{6-2 \cdot 3} \]
\[ = \binom{6}{3} (-1)^3 \frac{2^3}{3^3} \]

### Step 6: Calculate the Coefficient
Calculating \( \binom{6}{3} \):
\[ \binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]
Now substituting this value:
\[ T_{4} = 20 \cdot (-1) \cdot \frac{8}{27} = -\frac{160}{27} \]

### Final Answer
The term independent of \( x \) in the expansion of \( (2x - \frac{1}{3x})^6 \) is:
\[ -\frac{160}{27} \]