Key Concept: Binomial Theorem and Finding the Constant Term

This problem requires the application of the Binomial Theorem to find specific terms in an expansion. The general term, $T_{r+1}$, in the binomial expansion of $(a+b)^n$ is given by: $T_{r+1} = \binom{n}{r} a^{n-r} b^r$ To find the constant term (the term independent of $x$) in a product of expressions, we need to ensure that the powers of $x$ from each multiplied term sum to zero.

---

Step 1: Find the General Term of the Binomial Expansion

We begin by finding the general term of the binomial part of the expression, $\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9$. Here, $a = \frac{3}{2} x^2$, $b = -\frac{1}{3 x}$, and $n=9$.

Using the general term formula: $T_{r+1} = \binom{9}{r} \left(\frac{3}{2} x^2\right)^{9-r} \left(-\frac{1}{3 x}\right)^r$

Now, we separate the numerical coefficients and the powers of $x$: $T_{r+1} = \binom{9}{r} \left(\frac{3}{2}\right)^{9-r} (x^2)^{9-r} \left(-\frac{1}{3}\right)^r (x^{-1})^r$ $T_{r+1} = \binom{9}{r} \left(\frac{3^{9-r}}{2^{9-r}}\right) x^{18-2r} \left(\frac{(-1)^r}{3^r}\right) x^{-r}$ $T_{r+1} = \binom{9}{r} (-1)^r \frac{3^{9-r}}{3^r} \frac{1}{2^{9-r}} x^{18-2r-r}$ $T_{r+1} = \binom{9}{r} (-1)^r 3^{9-2r} 2^{r-9} x^{18-3r}$

This expression gives us the $(r+1)$-th term of the binomial expansion. The power of $x$ in this general term is $18-3r$.

---

Step 2: Identify Terms Contributing to the Constant Term in the Overall Product

The complete expression is $\left(1+2 x-3 x^3\right)\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9$. For the overall product to have a constant term, each term from the first polynomial $(1+2 x-3 x^3)$ must combine with a specific term from the binomial expansion $\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9$ such that the powers of $x$ cancel out.

Let $T_{r+1}$ be the general term of the binomial expansion found in Step 1.

Case 1: The term $1$ from $(1+2x-3x^3)$ To get a constant term when multiplying by $1$, we need the term independent of $x$ (i.e., $x^0$) from the binomial expansion. Set the power of $x$ in $T_{r+1}$ to $0$: $18-3r = 0 \Rightarrow 3r = 18 \Rightarrow r = 6$ Since $r=6$ is a valid integer between $0$ and $9$, this term exists. The constant term from the binomial expansion is $T_{6+1} = T_7$: $T_7 = \binom{9}{6} (-1)^6 3^{9-2(6)} 2^{6-9}$ $T_7 = \binom{9}{3} (1) 3^{9-12} 2^{-3}$ $T_7 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} \times 3^{-3} \times 2^{-3}$ $T_7 = 84 \times \frac{1}{27} \times \frac{1}{8} = \frac{84}{216}$ This term contributes $1 \times \frac{84}{216}$ to the total constant term.

Case 2: The term $2x$ from $(1+2x-3x^3)$ To get a constant term when multiplying by $2x$, we need a term with $x^{-1}$ from the binomial expansion. Set the power of $x$ in $T_{r+1}$ to $-1$: $18-3r = -1 \Rightarrow 3r = 19 \Rightarrow r = \frac{19}{3}$ Since $r$ must be an integer, there is no term with $x^{-1}$ in the binomial expansion. Therefore, this case contributes $0$ to the total constant term.

Case 3: The term $-3x^3$ from $(1+2x-3x^3)$ To get a constant term when multiplying by $-3x^3$, we need a term with $x^{-3}$ from the binomial expansion. Set the power of $x$ in $T_{r+1}$ to $-3$: $18-3r = -3 \Rightarrow 3r = 21 \Rightarrow r = 7$ Since $r=7$ is a valid integer between $0$ and $9$, this term exists. The term with $x^{-3}$ from the binomial expansion is $T_{7+1} = T_8$: $T_8 = \binom{9}{7} (-1)^7 3^{9-2(7)} 2^{7-9}$ $T_8 = \binom{9}{2} (-1) 3^{9-14} 2^{-2}$ $T_8 = \frac{9 \times 8}{2 \times 1} \times (-1) \times 3^{-5} \times 2^{-2}$ $T_8 = 36 \times (-1) \times \frac{1}{243} \times \frac{1}{4} = -\frac{36}{972} = -\frac{1}{27}$ This term contributes $-3x^3 \times (-\frac{1}{27}x^{-3}) = -3 \times (-\frac{1}{27}) = \frac{3}{27} = \frac{1}{9}$ to the total constant term.

---

Step 3: Calculate the Constant Term $p$

The total constant term, $p$, is the sum of the contributions from each case: $p = (\text{Contribution from Case 1}) + (\text{Contribution from Case 2}) + (\text{Contribution from Case 3})$ $p = \frac{84}{216} + 0 + \frac{1}{9}$ We can simplify $\frac{84}{216}$ by dividing both by $12$: $\frac{84 \div 12}{216 \div 12} = \frac{7}{18}$. So, $p = \frac{7}{18} + \frac{1}{9}$ To add these fractions, we find a common denominator, which is $18$: $p = \frac{7}{18} + \frac{1 \times 2}{9 \times 2} = \frac{7}{18} + \frac{2}{18}$ $p = \frac{7+2}{18} = \frac{9}{18} = \frac{1}{2}$

---

Step 4: Calculate $108p$

The problem asks for the value of $108p$. $108p = 108 \times \frac{1}{2}$ $108p = 54$

---

Tips and Common Mistakes:

• Checking $r$: Always ensure that the value of $r$ you obtain is a non-negative integer and is less than or equal to $n$. If $r$ is fractional or negative, it means no such term exists.
• Sign Errors: Be very careful with negative signs, especially when terms like $(-1)^r$ are involved.
• Combining Powers of $x$: Double-check the exponent calculations for $x$.
• Considering all parts: Don't forget to account for all terms in the first polynomial $(1+2 x-3 x^3)$. Each term must be multiplied by the appropriate term from the binomial expansion to yield a constant.

---

Summary and Key Takeaway: To find the constant term in an expression involving a product of a polynomial and a binomial expansion:

1. Determine the general term of the binomial expansion, separating the numerical coefficients and the variable terms.
2. For each term in the polynomial, identify the corresponding power of $x$ needed from the binomial expansion to result in a constant when multiplied.
3. Calculate $r$ for each required power of $x$. If $r$ is not a valid integer, that combination yields no constant term.
4. Calculate the specific terms for valid $r$ values and sum them up to get the total constant term. This methodical approach ensures all contributions to the constant term are accounted for. The final answer is $54$.