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

The correct option is C

$-18$

Explanation for the correct option.

$\begin{array}{rcl}x& =& \frac{-3+i\sqrt{3}}{2}\\ & =& \frac{-1+i\sqrt{3}}{2}-1\\ & =& \omega -1\left[\mathrm{As}\omega =\frac{-1+i\sqrt{3}}{2}\right]\end{array}$

So, $(x^2+3x{)}^2(x^2+3x+1)$ will be

$$\begin{gathered} ={\left[{\left(\omega -1\right)}^2+3\left(\omega -1\right)\right]}^2\left[{\left(\omega -1\right)}^2+3\left(\omega -1\right)+1\right] \\ ={\left[\left(\omega -1\right)(\omega -1+3)\right]}^2\left[{\omega }^2+1-2\omega +3\omega -3+1\right] \\ ={\left[\right(\omega -1\left)\right(\omega +2\left)\right]}^2[{\omega }^2+\omega -1]={\left[{\omega }^2+\omega -2\right]}^2\left[{\omega }^2+\omega -1\right] \\ ={[{\omega }^2+\omega +3-2-3]}^2[{\omega }^2+\omega +1-1-1] \\ ={[{\omega }^2+\omega +1-3]}^2[{\omega }^2+\omega +1-2] \\ ={\left(-3\right)}^2\left(-2\right)\left[As1+\omega +{\omega }^2=0\right] \\ =9\times (-2) \\ =-18 \end{gathered}$$

Hence, option C is correct.