In the expansion of `(x^(3) - (1)/(x^(2)))^(15)` , the constant term,is

Options

A. `""^(15)C_(6)`
B. 0
C. `-""^(15)C_(9)`
D. 1

Answer

To find the constant term in the expansion of \((x^3 - \frac{1}{x^2})^{15}\), we can follow these steps:

### Step 1: Identify the General Term
The general term in the binomial 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, simplify the general term:
\[ T_{r+1} = \binom{15}{r} (x^{3(15-r)}) \left(-1\right)^r \left(\frac{1}{x^{2r}}\right) \]
This simplifies to:
\[ T_{r+1} = \binom{15}{r} (-1)^r x^{45 - 3r - 2r} = \binom{15}{r} (-1)^r x^{45 - 5r} \]

### Step 3: Find the Constant Term
To find the constant term, we need the exponent of \(x\) to be zero:
\[ 45 - 5r = 0 \]
Solving for \(r\):
\[ 5r = 45 \implies r = 9 \]

### Step 4: Calculate the Constant Term
Now, substitute \(r = 9\) back into the general term:
\[ T_{10} = \binom{15}{9} (-1)^9 \]
This simplifies to:
\[ T_{10} = -\binom{15}{9} \]

### Step 5: Calculate \(\binom{15}{9}\)
Using the property of binomial coefficients:
\[ \binom{15}{9} = \binom{15}{6} \]
Calculating \(\binom{15}{6}\):
\[ \binom{15}{6} = \frac{15!}{6!(15-6)!} = \frac{15!}{6!9!} = \frac{15 \times 14 \times 13 \times 12 \times 11 \times 10}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 5005 \]

Thus, the constant term is:
\[ T_{10} = -5005 \]

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

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