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

We need to find the sum of the coefficients of $$x^3$$ and $$x^{-13}$$ in $$(1+x)(1-x)^2\left(1 + \frac{3}{x} + \frac{3}{x^2} + \frac{1}{x^3}\right)^5$$.

Recognizing the binomial pattern, we note that $$1 + \frac{3}{x} + \frac{3}{x^2} + \frac{1}{x^3} = \left(1 + \frac{1}{x}\right)^3 = \frac{(x+1)^3}{x^3}$$, so that $$\left(1 + \frac{3}{x} + \frac{3}{x^2} + \frac{1}{x^3}\right)^5 = \frac{(x+1)^{15}}{x^{15}}$$.

Thus the product becomes $$(1+x)(1-x)^2 \cdot \frac{(x+1)^{15}}{x^{15}} = \frac{(1+x)^{16}(1-x)^2}{x^{15}}$$. Let $$P(x) = (1+x)^{16}(1-x)^2$$; then the coefficient of $$x^k$$ in the original expression is the coefficient of $$x^{k+15}$$ in $$P(x)$$.

To find the coefficient of $$x^3$$, we look for $$[x^{18}]$$ in $$P(x) = (1+x)^{16}(1 - 2x + x^2)$$. Since $$(1+x)^{16}$$ has degree 16, the only way to obtain $$x^{18}$$ is via $$\binom{16}{16}x^{16} \times x^2 = x^{18}$$. Hence $$[x^{18}]P(x) = \binom{16}{16} \times 1 = 1$$.

For the coefficient of $$x^{-13}$$ we need $$[x^2]$$ in $$P(x)$$, which is given by $$[x^2]P(x) = \binom{16}{2}(1) + \binom{16}{1}(-2) + \binom{16}{0}(1) = 120 - 32 + 1 = 89$$.

Adding these results yields $$1 + 89 = \mathbf{90}$$.

The answer is $$\mathbf{90}$$.