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 \(108p\):

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