Find the term independent of `x` in the expansion of the following binomials:
(i) `(x-(1)/(x) )^(14)`

Answer

To find the term independent of \( x \) in the expansion of \( (x - \frac{1}{x})^{14} \), we will use the Binomial Theorem. The Binomial Theorem states that:

\[ (a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r \]

In our case, we can set \( a = x \) and \( b = -\frac{1}{x} \), and \( n = 14 \).

### Step 1: Write the general term
The general term \( T_{r+1} \) in the expansion is given by:

\[ T_{r+1} = \binom{14}{r} (x)^{14-r} \left(-\frac{1}{x}\right)^r \]

### Step 2: Simplify the general term
Now, we simplify the general term:

\[ T_{r+1} = \binom{14}{r} x^{14-r} \cdot \left(-1\right)^r \cdot \frac{1}{x^r} \]

This can be rewritten as:

\[ T_{r+1} = \binom{14}{r} (-1)^r x^{14-r-r} = \binom{14}{r} (-1)^r x^{14-2r} \]

### Step 3: Find the term independent of \( x \)
To find the term independent of \( x \), we need to set the exponent of \( x \) to zero:

\[ 14 - 2r = 0 \]

Solving for \( r \):

\[ 14 = 2r \implies r = 7 \]

### Step 4: Substitute \( r \) back into the general term
Now we substitute \( r = 7 \) back into the general term to find the independent term:

\[ T_{8} = \binom{14}{7} (-1)^7 \]

### Step 5: Calculate \( \binom{14}{7} \)
Now we calculate \( \binom{14}{7} \):

\[ \binom{14}{7} = \frac{14!}{7! \cdot 7!} = \frac{14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} \]

Calculating this step by step:

1. \( 14 \times 13 = 182 \)
2. \( 182 \times 12 = 2184 \)
3. \( 2184 \times 11 = 24024 \)
4. \( 24024 \times 10 = 240240 \)
5. \( 240240 \times 9 = 2162160 \)
6. \( 2162160 \times 8 = 17297280 \)

Now divide by \( 7! = 5040 \):

\[ \frac{17297280}{5040} = 3432 \]

### Step 6: Final calculation of the independent term
Thus, the independent term is:

\[ T_{8} = 3432 \cdot (-1)^7 = -3432 \]

### Final Answer
The term independent of \( x \) in the expansion of \( (x - \frac{1}{x})^{14} \) is:

\[ \boxed{-3432} \]