Question

If $x=\frac{-3+i\sqrt{3}}{2}$ is a complex number, then the value of ${(x^2+3x)}^2(x^2+3x+1)$ is

• A. $-\frac{9}{8}$
• B. $6$
• C. $-18$
• D. $36$

Answer 1

Answer: (C)

$-18$

Given, $x=\frac{-3+i\sqrt{3}}{2}$
$x=\frac{-3\pm i\sqrt{3}}{2}$
$\Rightarrow x=\frac{-1+i\sqrt{3}}{2}-1=\omega -1$
$\therefore {(x^2+3x)}^2(x^2+3x+1)$
$={[{(\omega -1)}^2+3]}^2[{(\omega -1)}^2+3(\omega -1)+1]$
$={({\omega }^2+\omega -2)}^2({\omega }^2+\omega -1)$
$[\because 1+\omega +{\omega }^2=0]$
$={(-1-2)}^2(-1-1)=-18$