If the constant term in the binomial expansion of `(x^2-1/x)^n ,n in N` is 15, then find the value of `n`.

Explanation

To find the value of \( n \) such that the constant term in the binomial expansion of \( (x^2 - \frac{1}{x})^n \) is 15, we can follow these steps:

### Step 1: Identify the General Term
The general term \( T_{r+1} \) 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, \( a = x^2 \) and \( b = -\frac{1}{x} \). Therefore, the general term becomes:
\[ T_{r+1} = \binom{n}{r} (x^2)^{n-r} \left(-\frac{1}{x}\right)^r \]
This simplifies to:
\[ T_{r+1} = \binom{n}{r} (x^{2(n-r)}) \left(-1\right)^r (x^{-r}) = \binom{n}{r} (-1)^r x^{2n - 2r - r} = \binom{n}{r} (-1)^r x^{2n - 3r} \]

### Step 2: Find the Constant Term
The constant term occurs when the exponent of \( x \) is zero:
\[ 2n - 3r = 0 \]
From this equation, we can express \( r \) in terms of \( n \):
\[ 2n = 3r \implies r = \frac{2n}{3} \]

### Step 3: Ensure \( r \) is an Integer
Since \( r \) must be a non-negative integer, \( 2n \) must be divisible by 3. This implies:
\[ n = \frac{3k}{2} \quad \text{for some integer } k \]
Thus, \( n \) must be a multiple of 3.

### Step 4: Substitute \( r \) into the Binomial Coefficient
Now we substitute \( r \) back into the general term to find the constant term:
\[ T_{r+1} = \binom{n}{\frac{2n}{3}} (-1)^{\frac{2n}{3}} \]
We know that this constant term is given to be 15:
\[ \left| \binom{n}{\frac{2n}{3}} \right| = 15 \]

### Step 5: Find Possible Values of \( n \)
We can try different values of \( n \) that are multiples of 3:
1. For \( n = 3 \):
\[ r = \frac{2 \cdot 3}{3} = 2 \implies T_{3} = \binom{3}{2} = 3 \quad \text{(not 15)} \]
2. For \( n = 6 \):
\[ r = \frac{2 \cdot 6}{3} = 4 \implies T_{5} = \binom{6}{4} = 15 \quad \text{(this works)} \]
3. For \( n = 9 \):
\[ r = \frac{2 \cdot 9}{3} = 6 \implies T_{7} = \binom{9}{6} = 84 \quad \text{(not 15)} \]

### Conclusion
The only value of \( n \) that gives a constant term of 15 is \( n = 6 \).

Thus, the value of \( n \) is:
\[ \boxed{6} \]