The middle term in the expansion of `(x^(3)-(1)/(x^(3)))^(10)` is-

Options

A. 252
B. -252
C. 210
D. -210

Answer

To find the middle term in the expansion of \((x^3 - \frac{1}{x^3})^{10}\), we can follow these steps:

### Step 1: Identify the total number of terms
In the binomial expansion of \((a + b)^n\), the number of terms is given by \(n + 1\). Here, \(n = 10\), so the total number of terms is:
\[ 10 + 1 = 11 \]

### Step 2: Determine the middle term
Since the total number of terms (11) is odd, the middle term will be the \(\left(\frac{n}{2} + 1\right)\)th term. Therefore, the middle term is:
\[ \text{Middle term} = T_{6} \]

### Step 3: Use the binomial theorem to find \(T_6\)
The general term in the binomial expansion is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
In our case, \(a = x^3\), \(b = -\frac{1}{x^3}\), and \(n = 10\). For \(T_6\), we have \(r = 5\):
\[ T_6 = \binom{10}{5} (x^3)^{10-5} \left(-\frac{1}{x^3}\right)^5 \]

### Step 4: Calculate \(T_6\)
Substituting the values:
\[ T_6 = \binom{10}{5} (x^3)^5 \left(-\frac{1}{x^3}\right)^5 \]
\[ = \binom{10}{5} (x^{15}) \left(-\frac{1}{x^{15}}\right) \]
\[ = \binom{10}{5} (-1) \]
\[ = -\binom{10}{5} \]

### Step 5: Calculate \(\binom{10}{5}\)
The value of \(\binom{10}{5}\) can be calculated as:
\[ \binom{10}{5} = \frac{10!}{5! \cdot 5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252 \]

### Step 6: Final result
Thus, the middle term \(T_6\) is:
\[ T_6 = -252 \]

### Summary
The middle term in the expansion of \((x^3 - \frac{1}{x^3})^{10}\) is \(-252\).

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