The coefficient of the term involving `x^2` in the expansion of `(3x - 1/x)^6` is 1210

Answer

To determine whether the coefficient of the term involving \(x^2\) in the expansion of \((3x - \frac{1}{x})^6\) is 1210, we will follow these steps:

### Step 1: Write the General Term
The general term \(T_r\) in the binomial expansion of \((a + b)^n\) is given by:
\[ T_r = \binom{n}{r} a^{n-r} b^r \]
For our expression \((3x - \frac{1}{x})^6\), we have:
- \(a = 3x\)
- \(b = -\frac{1}{x}\)
- \(n = 6\)

Thus, the general term becomes:
\[ T_r = \binom{6}{r} (3x)^{6-r} \left(-\frac{1}{x}\right)^r \]

### Step 2: Simplify the General Term
Now, we can simplify \(T_r\):
\[ T_r = \binom{6}{r} (3^{6-r} x^{6-r}) \left(-1^r \frac{1}{x^r}\right) = \binom{6}{r} 3^{6-r} (-1)^r x^{6-r - r} \]
This simplifies to:
\[ T_r = \binom{6}{r} 3^{6-r} (-1)^r x^{6-2r} \]

### Step 3: Find the Value of \(r\) for \(x^2\)
We want the term where the power of \(x\) is 2:
\[ 6 - 2r = 2 \]
Solving for \(r\):
\[ 6 - 2 = 2r \\ 4 = 2r \\ r = 2 \]

### Step 4: Calculate the Coefficient for \(r = 2\)
Now we substitute \(r = 2\) into the general term:
\[ T_2 = \binom{6}{2} 3^{6-2} (-1)^2 x^{6-2 \cdot 2} \]
Calculating each part:
- \(\binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15\)
- \(3^{6-2} = 3^4 = 81\)
- \((-1)^2 = 1\)

Thus, the coefficient becomes:
\[ \text{Coefficient} = 15 \cdot 81 = 1215 \]

### Conclusion
The coefficient of the term involving \(x^2\) in the expansion of \((3x - \frac{1}{x})^6\) is 1215, not 1210. Therefore, the statement is false.

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