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Step-by-Step Solution
Step 1: Identify Each Factor as a Binomial Expansion
We want the coefficient of x^{30} in the expansion of
(1 + \tfrac{1}{x})^6 (1 + x^2)^7 (1 - x^3)^8,
where x \neq 0. First, write each factor as a sum:
(1 + \tfrac{1}{x})^6 = \displaystyle \sum_{k=0}^{6} \binom{6}{k} x^{-k}.
(1 + x^2)^7 = \displaystyle \sum_{m=0}^{7} \binom{7}{m} x^{2m}.
(1 - x^3)^8 = \displaystyle \sum_{n=0}^{8} \binom{8}{n} (-1)^n x^{3n}.
Step 2: Combine Exponents in the Product
A typical term in the product after multiplying one term from each expansion is:
\binom{6}{k} x^{-k} \times \binom{7}{m} x^{2m} \times \binom{8}{n} (-1)^n x^{3n}.
The overall power of x in that term is
(-k) + (2m) + (3n).
We need this total to be 30:
-k + 2m + 3n = 30.
Step 3: Express k and Determine Valid Ranges
From the equation
-k + 2m + 3n = 30,
we get
k = 2m + 3n - 30.
Because k , m , and n must lie within their expansion limits:
0 \le k \le 6,\quad 0 \le m \le 7,\quad 0 \le n \le 8,
it follows that
0 \le 2m + 3n - 30 \le 6
\quad\Longrightarrow\quad
30 \le 2m + 3n \le 36.
Step 4: Find All Integer Solutions for (k, m, n)
We search for all integer pairs (m,n) such that 2m + 3n is between 30 and 36, inclusive. Then we set k = 2m + 3n - 30 . Check that k ends up between 0 and 6. The valid solutions are:
(k,m,n) = (0,6,6) , since 2\cdot6 + 3\cdot6 = 30.
(k,m,n) = (2,7,6) , since 2\cdot7 + 3\cdot6 = 32.
(k,m,n) = (1,5,7) , since 2\cdot5 + 3\cdot7 = 31.
(k,m,n) = (3,6,7) , since 2\cdot6 + 3\cdot7 = 33.
(k,m,n) = (5,7,7) , since 2\cdot7 + 3\cdot7 = 35.
(k,m,n) = (0,3,8) , since 2\cdot3 + 3\cdot8 = 30.
(k,m,n) = (2,4,8) , since 2\cdot4 + 3\cdot8 = 32.
(k,m,n) = (4,5,8) , since 2\cdot5 + 3\cdot8 = 34.
(k,m,n) = (6,6,8) , since 2\cdot6 + 3\cdot8 = 36.
Step 5: Compute the Contribution of Each Valid Triple
For each valid triple (k,m,n) , the product of binomial coefficients and the factor (-1)^n is:
\binom{6}{k} \times \binom{7}{m} \times \binom{8}{n} \times (-1)^n.
Let's list each:
(0,6,6) : \binom{6}{0}\,\binom{7}{6}\,\binom{8}{6}\,(-1)^6 = 1 \times 7 \times 28 \times 1 = 196.
(2,7,6) : \binom{6}{2}\,\binom{7}{7}\,\binom{8}{6}\,(-1)^6 = 15 \times 1 \times 28 \times 1 = 420.
(1,5,7) : \binom{6}{1}\,\binom{7}{5}\,\binom{8}{7}\,(-1)^7 = 6 \times 21 \times 8 \times (-1) = -1008.
(3,6,7) : \binom{6}{3}\,\binom{7}{6}\,\binom{8}{7}\,(-1)^7 = 20 \times 7 \times 8 \times (-1) = -1120.
(5,7,7) : \binom{6}{5}\,\binom{7}{7}\,\binom{8}{7}\,(-1)^7 = 6 \times 1 \times 8 \times (-1) = -48.
(0,3,8) : \binom{6}{0}\,\binom{7}{3}\,\binom{8}{8}\,(-1)^8 = 1 \times 35 \times 1 \times 1 = 35.
(2,4,8) : \binom{6}{2}\,\binom{7}{4}\,\binom{8}{8}\,(-1)^8 = 15 \times 35 \times 1 \times 1 = 525.
(4,5,8) : \binom{6}{4}\,\binom{7}{5}\,\binom{8}{8}\,(-1)^8 = 15 \times 21 \times 1 \times 1 = 315.
(6,6,8) : \binom{6}{6}\,\binom{7}{6}\,\binom{8}{8}\,(-1)^8 = 1 \times 7 \times 1 \times 1 = 7.
Step 6: Sum All Contributions
Sum of positive contributions: 196 + 420 + 35 + 525 + 315 + 7 = 1498.
Sum of negative contributions: (-1008) + (-1120) + (-48) = -2176.
Overall total: 1498 + (-2176) = -678.
Hence the coefficient of x^{30} is \alpha = -678. Therefore,
|\alpha| = 678.
Final Answer
\boxed{678}
