Find the term, which is independent of x in the expansion of
`(x^2 + 1/x)^9`

Answer

To find the term independent of \( x \) in the expansion of \( (x^2 + \frac{1}{x})^9 \), 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, \( a = x^2 \), \( b = \frac{1}{x} \), and \( n = 9 \). Thus, the general term becomes:
\[ T_{r+1} = \binom{9}{r} (x^2)^{9-r} \left(\frac{1}{x}\right)^r \]

### Step 2: Simplify the General Term
Now, we simplify \( T_{r+1} \):
\[ T_{r+1} = \binom{9}{r} (x^2)^{9-r} \cdot \frac{1}{x^r} = \binom{9}{r} x^{2(9-r)} \cdot x^{-r} \]
This can be rewritten as:
\[ T_{r+1} = \binom{9}{r} x^{18 - 2r - r} = \binom{9}{r} x^{18 - 3r} \]

### Step 3: Set the Exponent of \( x \) to Zero
To find the term that is independent of \( x \), we need to set the exponent of \( x \) to zero:
\[ 18 - 3r = 0 \]

### Step 4: Solve for \( r \)
Solving the equation:
\[ 18 = 3r \implies r = \frac{18}{3} = 6 \]

### Step 5: Find the Independent Term
Now, we substitute \( r = 6 \) back into the general term to find the specific term:
\[ T_{7} = \binom{9}{6} x^{18 - 3 \cdot 6} \]
Calculating \( T_{7} \):
\[ T_{7} = \binom{9}{6} x^{0} = \binom{9}{6} \]
Since \( \binom{9}{6} = \binom{9}{3} \):
\[ \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = \frac{504}{6} = 84 \]

### Final Answer
Thus, the term independent of \( x \) in the expansion of \( (x^2 + \frac{1}{x})^9 \) is:
\[ \boxed{84} \]