If the constant term in the expansion of left | vedclass

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Question

$\left(1+2 x-3 x^3\right)\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9$

General term $\mathrm{m}\left(\frac{3}{2} \mathrm{x}^2-\frac{1}{3 \mathrm{x}

Vedclass · Accepted Answer

$54$

Vedclass · Answer

$\left(1+2 x-3 x^3ight)\left(\frac{3}{2} x^2-\frac{1}{3 x}ight)^9$

General term $\mathrm{m}\left(\frac{3}{2} \mathrm{x}^2-\frac{1}{3 \mathrm{x}}ight)^9$

$={ }^9 \mathrm{C}_{\mathrm{r}} \cdot \frac{3^{9-2 \mathrm{r}}}{2^{9-\mathrm{r}}}(-1)^{\mathrm{r}} \cdot \mathrm{x}^{18-3 \mathrm{r}}$

Put $r=6$ to get coeff. of $x^0={ }^9 \mathrm{C}_6 \cdot \frac{1}{6^3} \cdot x^0=\frac{7}{18} x^0$

Put $r=7$ to get coeff. of $x^{-3}={ }^9 C_r \cdot \frac{3^{-5}}{2^2}(-1)^7 \cdot x^{-3}$

$ =-{ }^9 C_7 \cdot \frac{1}{3^5 \cdot 2^2} \cdot x^{-3}=\frac{-1}{27} x^{-3} $

$ \left(1+2 x-3 x^3ight)\left(\frac{7}{18} x^0-\frac{1}{27} x^{-3}ight) $

$ \frac{7}{18}+\frac{3}{27}=\frac{7}{18}+\frac{1}{9}=\frac{7+2}{18}=\frac{9}{18}=\frac{1}{2} $

$ 	herefore 108 \cdot \frac{1}{2}=54$

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

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