If the coefficient of $$x^{30}$$ in the expansion of $$\left(1 + \frac{1}{x}\right)^6 (1+x^2)^7 (1-x^3)^8$$; $$x \neq 0$$ is $$\alpha$$, then $$|\alpha|$$ equals:

We need to find the coefficient of $$x^{30}$$ in the expansion of $$\left(1 + \frac{1}{x}\right)^6(1 + x^2)^7(1 - x^3)^8$$, where $$x \neq 0$$.

Rewrite the factor $$\left(1 + \frac{1}{x}\right)^6$$ as $$\frac{(1+x)^6}{x^6}$$ so that the expression becomes $$\frac{(1+x)^6(1+x^2)^7(1-x^3)^8}{x^6}$$ and hence the coefficient of $$x^{30}$$ in the original product equals the coefficient of $$x^{36}$$ in $$(1+x)^6(1+x^2)^7(1-x^3)^8$$.

By the binomial theorem, $$(1+x)^6=\sum_{a=0}^6\binom{6}{a}x^a\,,\quad(1+x^2)^7=\sum_{b=0}^7\binom{7}{b}x^{2b}\,,\quad(1-x^3)^8=\sum_{c=0}^8\binom{8}{c}(-1)^cx^{3c}\,,$$ so we require $$a+2b+3c=36$$ with $$0\le a\le6\,,\;0\le b\le7\,,\;0\le c\le8\,. $$

Since $$a+2b\le6+14=20$$, it follows that $$3c=36-(a+2b)\ge36-20=16$$ so $$c\ge6$$, while $$3c\le36$$ gives $$c\le12$$ and hence $$6\le c\le8$$.

Case c = 6: Then $$a+2b=36-18=18$$ and writing $$a=18-2b$$ with $$0\le a\le6$$ and $$0\le b\le7$$ shows that $$b\ge6$$ and $$b\le7$$, yielding $$(b,a)=(6,6)\text{ or }(7,4)\,. $$

Case c = 7: Then $$a+2b=36-21=15$$ and $$a=15-2b$$ with the same bounds gives $$b\ge5$$ and $$b\le7$$, so $$(b,a)=(5,5),(6,3),(7,1)\,. $$

Case c = 8: Then $$a+2b=36-24=12$$ and $$a=12-2b$$ yields $$b\ge3$$ and $$b\le6$$, hence $$(b,a)=(3,6),(4,4),(5,2),(6,0)\,. $$

The contribution of each term is $$\binom{6}{a}\binom{7}{b}\binom{8}{c}(-1)^c\,. $$ For c = 6 one has $$(-1)^6=1$$ and $$\binom{8}{6}=28$$, giving contributions $$\binom{6}{6}\binom{7}{6}\cdot28=196$$ and $$\binom{6}{4}\binom{7}{7}\cdot28=420\,. $$

For c = 7 one has $$(-1)^7=-1$$ and $$\binom{8}{7}=8$$, giving contributions $$\binom{6}{5}\binom{7}{5}\cdot8\cdot(-1)=-1008\,,\quad\binom{6}{3}\binom{7}{6}\cdot8\cdot(-1)=-1120\,,\quad\binom{6}{1}\binom{7}{7}\cdot8\cdot(-1)=-48\,. $$

For c = 8 one has $$(-1)^8=1$$ and $$\binom{8}{8}=1$$, giving contributions $$\binom{6}{6}\binom{7}{3}=35\,,\;\binom{6}{4}\binom{7}{4}=525\,,\;\binom{6}{2}\binom{7}{5}=315\,,\;\binom{6}{0}\binom{7}{6}=7\,. $$

Summing all contributions yields $$\alpha=196+420-1008-1120-48+35+525+315+7=-678\,$$ and a quick check shows the positive terms sum to 1498 while the negative terms sum to 2176, giving $$1498-2176=-678\,. $$

Therefore the absolute value is $$|\alpha|=678$$ and the answer is 678.