Given expansion is (1+x1)6(1+x2)7(1−x3)8.
Now, coefficient of x30 in x6(x+1)6(1+x2)7(1−x3)8
So, we will find the coefficient of x36 in (1+x)6(1+x2)7(1−x3)8
General term of the given expansion is6Cr17Cr28Cr3(−1)r3xr1+2r2+3r3.
⇒r1+2r2+3r3=36
Case-I:
| r1 | r2 | r3 |
| 0 | 6 | 8 |
| 2 | 5 | 8 |
| 4 | 4 | 8 |
| 6 | 3 | 8 |
⇒r1+2r2=12 (Taking
r3=8)
Case II:
| r1 | r2 | r3 |
| 1 | 7 | 7 |
| 3 | 6 | 7 |
| 5 | 5 | 7 |
⇒r1+2r2=15 (Taking
r3=7)
Case III:
| r1 | r2 | r3 |
| 4 | 7 | 6 |
| 6 | 6 | 6 |
⇒r1+2r2=18 (Taking
r3=6)
So, the required coefficient Is given by, α=7+(15×21)+(15×35)+(35)−(6×8)−(20×7×8)−(6×21×8)+(15×28)+(7×28)
⇒α=−678
⇒∣α∣=678