Find the term independent of 'x' , `x ne 0` in the expansion of `(x^2 + 3/x)^6`

Answer

To find the term independent of 'x' in the expansion of \((x^2 + \frac{3}{x})^6\), 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:
\[ T_{r+1} = \binom{n}{r} a^{n-r} b^r \]
For our expression, \(a = x^2\) and \(b = \frac{3}{x}\), and \(n = 6\). Therefore, the general term becomes:
\[ T_{r+1} = \binom{6}{r} (x^2)^{6-r} \left(\frac{3}{x}\right)^r \]

### Step 2: Simplify the General Term
Now, simplify \(T_{r+1}\):
\[ T_{r+1} = \binom{6}{r} (x^2)^{6-r} \cdot \frac{3^r}{x^r} \]
This can be rewritten as:
\[ T_{r+1} = \binom{6}{r} 3^r x^{2(6-r)} x^{-r} = \binom{6}{r} 3^r x^{12 - 2r - r} = \binom{6}{r} 3^r x^{12 - 3r} \]

### Step 3: Find the Term Independent of 'x'
To find the term independent of \(x\), we need to set the exponent of \(x\) to zero:
\[ 12 - 3r = 0 \]
Solving for \(r\):
\[ 3r = 12 \implies r = 4 \]

### Step 4: Substitute \(r\) Back into the General Term
Now substitute \(r = 4\) into the general term:
\[ T_{5} = \binom{6}{4} 3^4 x^{12 - 3 \cdot 4} \]
Calculating \(T_{5}\):
\[ T_{5} = \binom{6}{4} 3^4 x^0 = \binom{6}{4} \cdot 81 \]
Since \(\binom{6}{4} = \binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15\), we have:
\[ T_{5} = 15 \cdot 81 = 1215 \]

### Final Answer
Thus, the term independent of \(x\) in the expansion of \((x^2 + \frac{3}{x})^6\) is:
\[ \boxed{1215} \]