Find the specified term of the expression in each of the following binomials:
(iv) Middle term of `(x^(4) - (1)/( x^3) )^(11)`.

Answer

To find the middle term of the expression \((x^4 - \frac{1}{x^3})^{11}\), we will follow these steps:

### Step 1: Identify the values of \(n\) and the terms in the binomial
The expression can be rewritten as:
\[ (x^4 + (-\frac{1}{x^3}))^{11} \]
Here, \(n = 11\), \(a = x^4\), and \(b = -\frac{1}{x^3}\).

### Step 2: Determine the middle term
For a binomial expansion, if \(n\) is odd, the middle terms are the \(\frac{n}{2} + 1\)th and \(\frac{n}{2} + 2\)th terms. Since \(n = 11\) (which is odd):
- The middle terms will be the 6th term and the 7th term.

### Step 3: Write the general term formula
The general term \(T_{r+1}\) in the binomial expansion is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
Substituting our values, we have:
\[ T_{r+1} = \binom{11}{r} (x^4)^{11-r} \left(-\frac{1}{x^3}\right)^r \]

### Step 4: Calculate the 6th term (\(T_6\))
For the 6th term, \(r = 5\):
\[ T_6 = \binom{11}{5} (x^4)^{11-5} \left(-\frac{1}{x^3}\right)^5 \]
Calculating each part:
- \(\binom{11}{5} = \frac{11!}{5!6!} = 462\)
- \((x^4)^{6} = x^{24}\)
- \(\left(-\frac{1}{x^3}\right)^5 = -\frac{1}{x^{15}}\)

Putting it all together:
\[ T_6 = 462 \cdot x^{24} \cdot \left(-\frac{1}{x^{15}}\right) = 462 \cdot -x^{24 - 15} = -462 x^9 \]

### Step 5: Calculate the 7th term (\(T_7\))
For the 7th term, \(r = 6\):
\[ T_7 = \binom{11}{6} (x^4)^{11-6} \left(-\frac{1}{x^3}\right)^6 \]
Calculating each part:
- \(\binom{11}{6} = \binom{11}{5} = 462\)
- \((x^4)^{5} = x^{20}\)
- \(\left(-\frac{1}{x^3}\right)^6 = \frac{1}{x^{18}}\)

Putting it all together:
\[ T_7 = 462 \cdot x^{20} \cdot \frac{1}{x^{18}} = 462 \cdot x^{20 - 18} = 462 x^2 \]

### Final Result
The middle terms of the expression \((x^4 - \frac{1}{x^3})^{11}\) are:
- 6th term: \(-462 x^9\)
- 7th term: \(462 x^2\)