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

- 9

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- 11

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Solution

The correct option is A $- 9$
The general term in the expansion of ${(1 + 3 x - 2 x^{2})}^{3}$ is:
$\frac{3!}{r_{1}! r_{2}! r_{3}!} \times {(1)}^{r_{1}} \times {(3 x)}^{r_{2}} \times {(- 2 x^{2})}^{r_{3}}$

$= \frac{3! \times {(3)}^{r_{2}} \times {(- 2)}^{r_{3}}}{r_{1}! r_{2}! r_{3}!} x^{r_{2} + 2 r_{3}}$

For coefficient of $x^{3}$ , it has to be $r_{2} + 2 r_{3} = 3$
There can be two possible combinations: $r_{1} = 1 r_{2} = 1 r_{3} = 1 r_{1} = 0 r_{2} = 3 r_{3} = 0$
So, the coefficient of $x^{3}$ will be
$\frac{{(- 2)}^{1} . 3^{1} .3!}{1! 1! 1!} + \frac{{(- 2)}^{0} . 3^{3} .3!}{3!}$
$= - 36 + 27 = - 9$

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