Find the constant term in expansion of `(x-(1)/(x))^(10)`

Answer

To find the constant term in the expansion of \((x - \frac{1}{x})^{10}\), we will use the Binomial Theorem. Let's go through the solution step by step.

### Step 1: Identify the general term in the expansion
According to 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, \(a = x\), \(b = -\frac{1}{x}\), and \(n = 10\). Therefore, the general term can be written as:
\[ T_{r+1} = \binom{10}{r} x^{10-r} \left(-\frac{1}{x}\right)^r \]

### Step 2: Simplify the general term
Now, simplifying the general term:
\[ T_{r+1} = \binom{10}{r} x^{10-r} \cdot (-1)^r \cdot \frac{1}{x^r} = \binom{10}{r} (-1)^r x^{10-r-r} = \binom{10}{r} (-1)^r x^{10-2r} \]

### Step 3: Find the constant term
To find the constant term, we need the exponent of \(x\) to be zero:
\[ 10 - 2r = 0 \]
Solving for \(r\):
\[ 2r = 10 \implies r = 5 \]

### Step 4: Identify the term corresponding to \(r = 5\)
The constant term corresponds to \(r = 5\). Thus, we need to find the 6th term (since \(r\) starts from 0):
\[ T_{6} = \binom{10}{5} (-1)^5 \]

### Step 5: Calculate the binomial coefficient and the constant term
Calculating \(\binom{10}{5}\):
\[ \binom{10}{5} = \frac{10!}{5!5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252 \]
Now substituting back into the term:
\[ T_{6} = 252 \cdot (-1)^5 = 252 \cdot (-1) = -252 \]

### Final Answer
The constant term in the expansion of \((x - \frac{1}{x})^{10}\) is \(-252\).

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