Consider the expansion `(x^(3)-(1)/(x^2))^(15)`.
What is the independent term in the given expansion ?

Options

A. `""^(15)C_(5)`
B. 0
C. `-""^(15)C_(9)`
D. 1

Explanation

To find the independent term in the expansion of \((x^3 - \frac{1}{x^2})^{15}\), we will use the binomial theorem. Here are the steps to solve the problem:

### Step 1: Identify the general term in the expansion
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, \(a = x^3\), \(b = -\frac{1}{x^2}\), and \(n = 15\). Thus, the general term becomes:
\[ T_{r+1} = \binom{15}{r} (x^3)^{15-r} \left(-\frac{1}{x^2}\right)^r \]

### Step 2: Simplify the general term
Now, we can simplify \(T_{r+1}\):
\[ T_{r+1} = \binom{15}{r} (x^{3(15-r)}) \left(-\frac{1}{x^{2r}}\right) \]
This can be rewritten as:
\[ T_{r+1} = \binom{15}{r} (-1)^r x^{45 - 3r - 2r} = \binom{15}{r} (-1)^r x^{45 - 5r} \]

### Step 3: Find the condition for the independent term
The independent term occurs when the exponent of \(x\) is zero:
\[ 45 - 5r = 0 \]
Solving for \(r\):
\[ 5r = 45 \implies r = 9 \]

### Step 4: Determine the independent term
Now that we have \(r = 9\), we can find the independent term by substituting \(r\) back into the general term:
\[ T_{10} = \binom{15}{9} (-1)^9 \]
Calculating \(\binom{15}{9}\):
\[ \binom{15}{9} = \binom{15}{6} = \frac{15!}{9!6!} = 5005 \]
Thus,
\[ T_{10} = 5005 \cdot (-1) = -5005 \]

### Conclusion
The independent term in the expansion of \((x^3 - \frac{1}{x^2})^{15}\) is \(-5005\).

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