The question is asking to determine the coefficient of the constant term in the binomial expansion of the expression:

$\left( x^3 - \frac{3}{x^2} \right)^5.$

Step-by-step Solution:

1. General Binomial Expansion: The binomial theorem for expanding $(a + b)^n$ is:

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

   In this case, we have $a = x^3$ and $b = -\frac{3}{x^2}$, and $n = 5$.

2. Find Constant Term: A constant term will be found when the powers of $x$ in both terms cancel out, i.e., when the total exponent of $x$ becomes zero.

   We need to find terms in the expansion where:

   $(x^3)^{5-k} \left(-\frac{3}{x^2}\right)^k$

   results in no $x$-terms, meaning:

   $3(5-k) - 2k = 0.$

3. Solve for $k$: $15 - 3k - 2k = 0 \implies 15 = 5k \implies k = 3.$

4. Find the Corresponding Coefficient: For $k = 3$, the corresponding term in the expansion is:

   $\binom{5}{3} (x^3)^{5-3} \left(-\frac{3}{x^2}\right)^3.$

   Simplifying this:

   $\binom{5}{3} (x^3)^2 \left(-\frac{3}{x^2}\right)^3 = \binom{5}{3} x^6 \left(-\frac{27}{x^6}\right).$

   The powers of $x$ cancel out, leaving:

   $\binom{5}{3} (-27).$

   The binomial coefficient $\binom{5}{3} = 10$, so the coefficient of the constant term is:

   $10 \times (-27) = -270.$

Final Answer:

The coefficient of the constant term is -270.

Would you like any more details or explanations about this solution?

Related Questions:

1. How does the binomial theorem work for higher powers?
2. How do you compute binomial coefficients?
3. What are the general conditions for finding constant terms in a binomial expansion?
4. Can we expand the binomial for fractional exponents?
5. How would the expansion change if we had a different exponent (e.g., 6 instead of 5)?

Tip:

When looking for constant terms in binomial expansions, always focus on how the powers of variables cancel out.