If the constant term in the expansion of (1+2 x-3 x^3)((3 / 2) x^2-(1 / 3 x))^9 is p, then 108 p is equal to ________.

General term of ((3 / 2) x^2-(1 / 3 x))^9 T_r+1= ^9 C_r((3 / 2) x^2)^9-r(-(1 / 3 x))^r \ = ^9 C_r(-1)^r 3^9-2 r 2^r-9 x^18-35 Constant term in expansion of (1+2 x-3 x^3) ((3 / 2) x^2-(1 / 3 x))^9 \ =T_7-3 T_8= ^9 C_6 3^-3 * 2^-3+3 ^9 C_7 * 3^-5 * 2^-2 \ =(3 x 4 x 7 / 3^3 * 2^3)+(3 x 9 x 4 / 3^5 x 2^2)= \ p=(42+12 / 108)=(54 / 108) \ 108 p=54 \

If the constant term in the expansion of 12 x3 x3frac 3... | null (Mathematics)

\text { General term of }\left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9

T_{r+1}={ }^9 C_r\left(\frac{3}{2} x^2\right)^{9-r}\left(-\frac{1}{3 x}\right)^r \\\\ ={ }^9 C_r(-1)^r 3^{9-2 r} 2^{r-9} x^{18-35}

Constant term in expansion of $\left(1+2 x-3 x^3\right)$

\begin{aligned} & \left(\frac{3}{2} x^2-\frac{1}{3 x}\right)^9 \\ & =T_7-3 T_8={ }^9 C_6 3^{-3} \cdot 2^{-3}+3{ }^9 C_7 \cdot 3^{-5} \cdot 2^{-2} \\ & =\frac{3 \times 4 \times 7}{3^3 \cdot 2^3}+\frac{3 \times 9 \times 4}{3^5 \times 2^2}= \\ & p=\frac{42+12}{108}=\frac{54}{108} \\ & 108 p=54 \end{aligned}

Explanation

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