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!