Find the term independent of x in `((sqrtx)/(3) - (4)/(xsqrtx))^12`

Answer

To find the term independent of \( x \) in the expression \(\left(\frac{\sqrt{x}}{3} - \frac{4}{x\sqrt{x}}\right)^{12}\), we will use the Binomial Theorem. Let's go through the steps:

### Step 1: Identify the General Term
Using the Binomial Theorem, the general term \( T_{r+1} \) in the expansion of \( (a + b)^n \) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
In our case, \( n = 12 \), \( a = \frac{\sqrt{x}}{3} \), and \( b = -\frac{4}{x\sqrt{x}} \).

### Step 2: Write the General Term
Substituting the values into the formula, we have:
\[ T_{r+1} = \binom{12}{r} \left(\frac{\sqrt{x}}{3}\right)^{12-r} \left(-\frac{4}{x\sqrt{x}}\right)^{r} \]

### Step 3: Simplify the General Term
Now, we simplify the expression:
\[ T_{r+1} = \binom{12}{r} \left(\frac{x^{1/2}}{3}\right)^{12-r} \left(-\frac{4}{x^{3/2}}\right)^{r} \]
This can be rewritten as:
\[ T_{r+1} = \binom{12}{r} \frac{(-4)^r}{3^{12-r}} x^{\frac{12-r}{2} - \frac{3r}{2}} \]
\[ = \binom{12}{r} \frac{(-4)^r}{3^{12-r}} x^{\frac{12 - 4r}{2}} \]

### Step 4: Find the Power of \( x \)
For the term to be independent of \( x \), the exponent of \( x \) must be zero:
\[ \frac{12 - 4r}{2} = 0 \]
Multiplying through by 2 gives:
\[ 12 - 4r = 0 \]
Solving for \( r \):
\[ 4r = 12 \quad \Rightarrow \quad r = 3 \]

### Step 5: Substitute \( r \) Back into the General Term
Now, we substitute \( r = 3 \) back into the general term:
\[ T_{4} = \binom{12}{3} \frac{(-4)^3}{3^{12-3}} = \binom{12}{3} \frac{-64}{3^9} \]

### Step 6: Calculate \( \binom{12}{3} \)
Calculating \( \binom{12}{3} \):
\[ \binom{12}{3} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220 \]

### Step 7: Final Calculation
Now substituting this value back:
\[ T_{4} = 220 \cdot \frac{-64}{3^9} \]
Calculating \( 3^9 = 19683 \):
\[ T_{4} = \frac{-220 \cdot 64}{19683} = \frac{-14080}{19683} \]

### Conclusion
The term independent of \( x \) in the expression is:
\[ \frac{-14080}{19683} \]