The coefficient of `x^(3)` in the expansion of `(x-(1)/(x))^(7)` is:

Options

A. 14
B. 21
C. 28
D. 35

Answer

To find the coefficient of \( x^3 \) in the expansion of \( (x - \frac{1}{x})^7 \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ (x - \frac{1}{x})^7 \]

### Step 2: Use the Binomial Theorem
According to the binomial theorem, the expansion of \( (a + b)^n \) is given by:
\[ \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r \]
In our case, let \( a = x \) and \( b = -\frac{1}{x} \). Thus, we can write:
\[ (x - \frac{1}{x})^7 = \sum_{r=0}^{7} \binom{7}{r} x^{7-r} \left(-\frac{1}{x}\right)^r \]

### Step 3: Simplify the general term
The general term of the expansion becomes:
\[ \binom{7}{r} x^{7-r} \left(-1\right)^r \frac{1}{x^r} = \binom{7}{r} (-1)^r x^{7 - r - r} = \binom{7}{r} (-1)^r x^{7 - 2r} \]

### Step 4: Find the term where the power of \( x \) is 3
We need to find \( r \) such that:
\[ 7 - 2r = 3 \]
Solving for \( r \):
\[ 7 - 2r = 3 \implies 2r = 4 \implies r = 2 \]

### Step 5: Substitute \( r \) back into the general term
Now, substitute \( r = 2 \) into the general term:
\[ \text{Term} = \binom{7}{2} (-1)^2 x^{7 - 2 \cdot 2} = \binom{7}{2} x^3 \]

### Step 6: Calculate \( \binom{7}{2} \)
Now calculate \( \binom{7}{2} \):
\[ \binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7 \times 6}{2 \times 1} = 21 \]

### Step 7: Conclusion
The coefficient of \( x^3 \) in the expansion is:
\[ \text{Coefficient} = 21 \]