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The problem asks for the constant term in the expansion of \( \left( 3x^2 - \frac{1}{x} \right)^6 \).

We can solve this using the **Binomial Theorem**.

### Step 1: Binomial Expansion
The binomial expansion of \( (a + b)^n \) is given by:

\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\]

In this case, let:
- \( a = 3x^2 \)
- \( b = -\frac{1}{x} \)
- \( n = 6 \)

Thus, the expansion of \( \left( 3x^2 - \frac{1}{x} \right)^6 \) is:

\[
\sum_{k=0}^{6} \binom{6}{k} (3x^2)^{6-k} \left( -\frac{1}{x} \right)^k
\]

### Step 2: Simplifying the General Term
The general term in this expansion is:

\[
T_k = \binom{6}{k} (3x^2)^{6-k} \left( -\frac{1}{x} \right)^k
\]

Simplify each part:
\[
T_k = \binom{6}{k} 3^{6-k} x^{2(6-k)} (-1)^k x^{-k}
\]

Combine the powers of \(x\):
\[
T_k = \binom{6}{k} 3^{6-k} (-1)^k x^{12 - 3k}
\]

### Step 3: Find the Constant Term
For the term to be a constant, the power of \(x\) must be zero. Therefore, set the exponent of \(x\) equal to zero:
\[
12 - 3k = 0
\]

Solve for \(k\):
\[
k = 4
\]

### Step 4: Calculate the Constant Term
Substitute \(k = 4\) into the expression for the general term:

\[
T_4 = \binom{6}{4} 3^{6-4} (-1)^4 x^{12 - 3(4)}
\]

Since \( (-1)^4 = 1 \) and the power of \(x\) is zero (constant term), we now calculate:

\[
T_4 = \binom{6}{4} 3^2 = \binom{6}{4} 9
\]

The binomial coefficient \( \binom{6}{4} = \binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15 \).

Thus, the constant term is:
\[
T_4 = 15 \times 9 = 135
\]

### Final Answer:
The constant term in the expansion is **135**, which corresponds to option (b).

Would you like more details or have any questions?

Here are five related questions for further exploration:
1. What is the general form of the binomial expansion?
2. How do we find specific terms in a binomial expansion?
3. How can we apply this process to different powers in binomial expansions?
4. How does changing the coefficients affect the constant term in the expansion?
5. What is the role of negative exponents in binomial expansions?

**Tip:** Always check the powers of variables carefully when looking for a constant term in binomial expansions!

The constant term in the expansion of (3x^2 - 1/x)^6 is?

This solution walks you through how to find the constant term in the expansion of (3x^2 - 1/x)^6 using the binomial theorem. A step-by-step breakdown of how to identify the constant term by applying the binomial expansion formula. Suitable for high school or early college students.

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