$\begin{aligned} & (1+x)\left(1-x^2\right)\left(1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3}\right)^5 \\ & =(1+x)\left(1-x^2\right)\left(\left(1+\frac{1}{x}\right)^3\right)^5 \\ & =\frac{(1+x)^2(1-x)(1+x)^{15}}{x^{15}} \\ & =\frac{(1+x)^{17}-x(1+x)^{17}}{x^{15}} \\ & =\operatorname{coeff}\left(x^3\right) \text { in the expansion } \approx \operatorname{coeff}\left(x^{18}\right) \text { in } \\ & (1+x)^{17}-x(1+x)^{17} \\ & =0-1 \\ & =-1 \\ & \text { coeff }\left(x^{-13}\right) \text { in the expansion } \approx \operatorname{coeff}\left(x^2\right) \quad \text { in }\end{aligned}$

$(1+x)^{17}-x(1+x)^{17}$

$=0-1$

$=-1$

$\operatorname{coeff}\left(\mathrm{x}^{-13}\right)$ in the expansion $\approx \operatorname{coeff}\left(\mathrm{x}^2\right)$ in

$(1+x)^{17}-x(1+x)^{17}$

$=\left(\begin{array}{c}17 \\ 2\end{array}\right)-\left(\begin{array}{c}17 \\ 1\end{array}\right)$

$=17 \times 8-17$

$=17 \times 7$

$=119$

Hence Answer $=119-1=118$