Question

Find the coefficient of $x^5$ in the expansion of $(1+3x-2x^3{)}^8$

Answer 1

$(1+30x-2x^3{)}^8$
$=\sum \frac{n!}{(p!)(q!)(r!)}\left(1{)}^6\right(3x{)}^q(-2x^3{)}^r$

where p + q + r = n = 8

for $x^5$, we have q + 3r = 5

$\therefore$ r = 0, q = 5 $\therefore$ p = 3 by (I)
or r = 1, q = 2 $\therefore$ p = 5 by (I)

Putting the above values in (A), we have

$\frac{8!}{(3!)(5!)(0!)}{.1}^3{.3}^5(-2{)}^0+\frac{(8!)}{(5!)(2!)(1!)}{.1}^5{.3}^2.(-2{)}^1$

$=56\left(243\right)+\mathrm{8.7.6}(-9)=56(243-54)=56\left(189\right)$