If the coefficient of x ^ 30 in the expansion of (1 + (1)/(x)) ^ 6...

Given expansion is ${\left(1 + \dfrac{1}{x}\right)}^{6}{\left(1 + {x}^{2}\right)}^{7}{\left(1 - {x}^{3}\right)}^{8}$.

Now, coefficient of ${x}^{30}$ in $\dfrac{{\left(x + 1\right)}^{6}{\left(1 + {x}^{2}\right)}^{7}{\left(1 - {x}^{3}\right)}^{8}}{{x}^{6}}$

So, we will find the coefficient of ${x}^{36}$ in ${\left(1 + x\right)}^{6}{\left(1 + {x}^{2}\right)}^{7}{\left(1 - {x}^{3}\right)}^{8}$

General term of the given expansion is${}^{6}{{C}_{{r}_{1}}}^{7}{{C}_{{r}_{2}}}^{8}{C}_{{r}_{3}}{\left(- 1\right)}^{{r}_{3}}{x}^{{r}_{1} + 2{r}_{2} + 3{r}_{3}}$.

$⇒ {r}_{1} + 2{r}_{2} + 3{r}_{3} = 36$

Case-I:

$⇒ {r}_{1} + 2{r}_{2} = 12$ (Taking ${r}_{3} = 8$)

Case II:

$⇒ {r}_{1} + 2{r}_{2} = 15$ (Taking ${r}_{3} = 7$)

Case III:

$⇒ {r}_{1} + 2{r}_{2} = 18$ (Taking ${r}_{3} = 6$)

So, the required coefficient Is given by, $\alpha = 7 + \left(15 \times 21\right) + \left(15 \times 35\right) + \left(35\right) - \left(6 \times 8\right) - \left(20 \times 7 \times 8\right) - \left(6 \times 21 \times 8\right) + \left(15 \times 28\right) + \left(7 \times 28\right)$

$⇒ \alpha = - 678$

$⇒ \left|\alpha\right| = 678$