Question

If the constant term in the expansion of \((1 + 2x - 3x^3) \left(\frac{3}{2}x^2 - \frac{1}{3x}\right)^9\) is \( p \), then \( 108p \) is equal to _________.

Answer 1

Correct Answer: 54

Given expression:

\[ \left(1 + 2x - 3x^3\right)\left(\frac{3}{2}x^2 - \frac{1}{3x}\right)^9 \]

General term:

\[ T_r = \binom{9}{r} \left(\frac{3}{2}x^2\right)^{9-r} \left(-\frac{1}{3x}\right)^r \]

Simplifying:

\[ T_r = \binom{9}{r} \left(\frac{3}{2}\right)^{9-r} \left(-\frac{1}{3}\right)^r x^{2(9-r)}x^{-r} = \binom{9}{r} \left(\frac{3}{2}\right)^{9-r} \left(-\frac{1}{3}\right)^r x^{18-3r} \]

To find the constant term, we set the exponent of \(x\) to zero:

\[ 18 - 3r = 0 \implies r = 6 \]

Substituting \(r = 6\) into the expression:

\[ T_6 = \binom{9}{6} \left(\frac{3}{2}\right)^3 \left(-\frac{1}{3}\right)^6 \]

Calculating each term:

\[ \binom{9}{6} = 84, \quad \left(\frac{3}{2}\right)^3 = \frac{27}{8}, \quad \left(-\frac{1}{3}\right)^6 = \frac{1}{729} \] \[ T_6 = 84 \times \frac{27}{8} \times \frac{1}{729} = \frac{7}{18} \]

Next, substituting \(r = 7\) to find the coefficient of \(x^{-3}\):

\[ T_7 = \binom{9}{7} \left(\frac{3}{2}\right)^2 \left(-\frac{1}{3}\right)^7 \]

Calculating:

\[ \binom{9}{7} = 36, \quad \left(\frac{3}{2}\right)^2 = \frac{9}{4}, \quad \left(-\frac{1}{3}\right)^7 = -\frac{1}{2187} \] \[ T_7 = 36 \times \frac{9}{4} \times -\frac{1}{2187} = -\frac{1}{27} \]

Combining the terms:

\[ \left(1 + 2x - 3x^3\right)\left(\frac{7}{18} + \frac{-1}{27}x^3\right) \]

Simplifying:

\[ \text{Constant term} = \frac{7}{18} \]

Given that \(p\) is the constant term, we have \(p = \frac{7}{18}\). Calculating \(108p\):

\[ 108p = 108 \times \frac{7}{18} = 54 \]

Answer 2

Step 1: Restructure the expression and goal
We want the constant term (coefficient of \(x^0\)) in \[ (1 + 2x - 3x^3)\left(\frac{3}{2}x^2 - \frac{1}{3x}\right)^9. \] Let \[ A(x)=\left(\frac{3}{2}x^2 - \frac{1}{3x}\right)^9. \] We will expand \(A(x)\) using the binomial theorem, identify the powers of \(x\), and then select those that, when multiplied by \(1\), \(2x\), and \(-3x^3\), yield an overall power \(x^0\).

Step 2: Binomial expansion of \(A(x)\) and general term
Using the binomial theorem, \[ A(x)=\sum_{k=0}^{9} \binom{9}{k}\left(\frac{3}{2}x^2\right)^{9-k}\left(-\frac{1}{3x}\right)^k. \] The power of \(x\) in the \(k\)-th term is \[ x^{\,2(9-k)}\cdot x^{-k}=x^{\,18-3k}, \] and its coefficient is \[ \binom{9}{k}\left(\frac{3}{2}\right)^{9-k}\left(-\frac{1}{3}\right)^k =\binom{9}{k}\,(-1)^k\,\frac{3^{\,9-2k}}{2^{\,9-k}}. \] Therefore, \[ A(x)=\sum_{k=0}^{9} a_k\,x^{\,18-3k}\quad \text{with}\quad a_k=\binom{9}{k}\,(-1)^k\,\frac{3^{\,9-2k}}{2^{\,9-k}}. \]

Step 3: Which terms contribute to the constant term after multiplying by \((1+2x-3x^3)\)?
We need overall power \(x^0\). Let the term from \(A(x)\) be \(a_k x^{18-3k}\). Then:
1) Multiplying by \(1\) requires \(18-3k=0 \Rightarrow k=6\).
2) Multiplying by \(2x\) requires \(18-3k+1=0 \Rightarrow 3k=19\) (no integer solution).
3) Multiplying by \(-3x^3\) requires \(18-3k+3=0 \Rightarrow 3k=21 \Rightarrow k=7\).
Hence, only the \(k=6\) term (from the factor \(1\)) and the \(k=7\) term (from the factor \(-3x^3\)) contribute. There is no contribution from the \(2x\) term.

Step 4: Compute the needed coefficients \(a_6\) and \(a_7\)
For \(k=6\): \[ a_6=\binom{9}{6}(-1)^6\frac{3^{\,9-12}}{2^{\,9-6}} =\binom{9}{6}\cdot\frac{3^{-3}}{2^{3}} =\frac{\binom{9}{6}}{27\cdot 8}. \] Since \(\binom{9}{6}=\binom{9}{3}=84\), \[ a_6=\frac{84}{216}=\frac{7}{18}. \] For \(k=7\): \[ a_7=\binom{9}{7}(-1)^7\frac{3^{\,9-14}}{2^{\,9-7}} =-\,\binom{9}{7}\cdot\frac{3^{-5}}{2^{2}} =-\,\frac{\binom{9}{7}}{243\cdot 4}. \] Since \(\binom{9}{7}=\binom{9}{2}=36\), \[ a_7=-\frac{36}{972}=-\frac{1}{27}. \]

Step 5: Assemble the constant term \(p\)
Only \(1\cdot a_6 x^{0}\) and \((-3x^3)\cdot a_7 x^{-3}\) contribute: \[ p = a_6 + (-3)\,a_7 = \frac{7}{18} - 3\left(-\frac{1}{27}\right) = \frac{7}{18} + \frac{1}{9} = \frac{7}{18} + \frac{2}{18} = \frac{9}{18} = \frac{1}{2}. \]

Step 6: Compute \(108p\)
\[ 108p=108\cdot\frac{1}{2}=54. \]

Final answer

54