Sum of the coefficients of `x^(3)` and `x^(6)` in the expansion of `(x^(2)-1/x)^(9)` is

Answer

To find the sum of the coefficients of \(x^3\) and \(x^6\) in the expansion of \((x^2 - \frac{1}{x})^9\), we can follow these steps:

### Step 1: Identify the general term in the binomial expansion
The binomial expansion of \((a + b)^n\) is given by:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
In our case, \(a = x^2\), \(b = -\frac{1}{x}\), and \(n = 9\). Therefore, the general term \(T_{r+1}\) in the expansion is:
\[ T_{r+1} = \binom{9}{r} (x^2)^{9-r} \left(-\frac{1}{x}\right)^r \]
This simplifies to:
\[ T_{r+1} = \binom{9}{r} x^{2(9-r)} (-1)^r x^{-r} = \binom{9}{r} (-1)^r x^{18 - 3r} \]

### Step 2: Find the coefficient of \(x^3\)
To find the coefficient of \(x^3\), we set the exponent equal to 3:
\[ 18 - 3r = 3 \]
Solving for \(r\):
\[ 18 - 3r = 3 \implies 3r = 15 \implies r = 5 \]
Now, substitute \(r = 5\) into the general term to find the coefficient:
\[ \text{Coefficient of } x^3 = \binom{9}{5} (-1)^5 = -\binom{9}{5} \]

### Step 3: Find the coefficient of \(x^6\)
Next, we find the coefficient of \(x^6\) by setting the exponent equal to 6:
\[ 18 - 3r = 6 \]
Solving for \(r\):
\[ 18 - 3r = 6 \implies 3r = 12 \implies r = 4 \]
Now, substitute \(r = 4\) into the general term to find the coefficient:
\[ \text{Coefficient of } x^6 = \binom{9}{4} (-1)^4 = \binom{9}{4} \]

### Step 4: Sum the coefficients
Now we need to sum the coefficients of \(x^3\) and \(x^6\):
\[ \text{Sum} = \text{Coefficient of } x^3 + \text{Coefficient of } x^6 = -\binom{9}{5} + \binom{9}{4} \]
Using the property of binomial coefficients, we know:
\[ \binom{9}{5} = \binom{9}{4} \]
Thus, we can rewrite the sum as:
\[ \text{Sum} = -\binom{9}{4} + \binom{9}{4} = 0 \]

### Final Answer
The sum of the coefficients of \(x^3\) and \(x^6\) in the expansion of \((x^2 - \frac{1}{x})^9\) is:
\[ \boxed{0} \]