In the expansion of $$(1+x)\left(1-x^2\right)\left(1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3}\right)^5, x \neq 0$$, the sum of the coefficients of $x^3$ and $x^{-13}$ is equal to __________.

<p>$$\begin{aligned} & (1+x)\left(1-x^2\right)\left(1+\frac{3}{x}+\frac{3}{x^2}+\frac{1}{x^3}\right)^5 \\ & =(1+x)\left(1-x^2\right)\left(\left(1+\frac{1}{x}\right)^3\right)^5 \\ & =\frac{(1+x)^2(1-x)(1+x)^{15}}{x^{15}} \\ & =\frac{(1+x)^{17}-x(1+x)^{17}}{x^{15}} \end{aligned}$$</p> <p>$=\operatorname{coeff}\left(\mathrm{x}^3\right)$ in the expansion $\approx \operatorname{coeff}\left(\mathrm{x}^{18}\right)$ in</p> <p>$$\begin{aligned} & (1+x)^{17}-x(1+x)^{17} \\ & =0-1 \\ & =-1 \end{aligned}$$</p> <p>$\operatorname{coeff}\left(\mathrm{x}^{-13}\right)$ in the expansion $\approx \operatorname{coeff}\left(\mathrm{x}^2\right)$ in</p> <p>$$\begin{aligned} & (1+x)^{17}-x(1+x)^{17} \\ & =\left(\begin{array}{c} 17 \\ 2 \end{array}\right)-\left(\begin{array}{c} 17 \\ 1 \end{array}\right) \\ & =17 \times 8-17 \\ & =17 \times 7 \\ & =119 \end{aligned}$$</p> <p>Hence Answer $=119-1=118$</p>