Question

If $1,$ $\omega and{\omega }^2$ are the cube roots of unity, then $\omega \left({\left(1+\omega \right)}^3-{\left(1+\omega \right)}^2\right)$is equal to

• A. $1$
• B. $-1$
• C. $i$
• D. $0$

Answer 1

The correct option is A

$1$

Explanation for the correct option:

Step1. Define the cube root of unity:

Given,

$1,\omega and{\omega }^2$are the cube root of unity.

$$\begin{gathered} 1+\omega +{\omega }^2=0\_\_\_\left(1\right)\left[\mathbf{\because }\mathbf{Using}\mathbf{}\mathbf{definition}\mathbf{,}\mathbf{}\mathbf{}\mathbf{\omega }\mathbf{=}\frac{\mathbf{-}\mathbf{1}\mathbf{+}\mathbf{i}\sqrt{\mathbf{3}}}{\mathbf{2}}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{\omega }}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{-}\mathbf{1}\mathbf{-}\mathbf{i}\sqrt{\mathbf{3}}}{\mathbf{2}}\right] \\ 1\times \mathrm{\omega }\times {\mathrm{\omega }}^2=1\mathrm{\_\_\_\_}\left(2\right) \end{gathered}$$

Step2. Find the value of $\mathbf{\omega }\left({\left(1+\omega \right)}^3-{\left(1+\omega \right)}^2\right)$:

$$\begin{gathered} \omega \left({\left(1+\omega \right)}^3-{\left(1+\omega \right)}^2\right) \\ =\omega \left({\left(-{\omega }^2\right)}^3-{\left(-{\omega }^2\right)}^2\right)\left[\because 1+\omega =-{\omega }^2\right] \\ =\omega \left[{\left(-1\right)}^3{\left({\omega }^3\right)}^2-\left({\omega }^3\right)\left(\omega \right)\right] \\ =\omega \left[-1{\left(1\right)}^2-\left(1\right)\left(\omega \right)\right]\left[\because From\left(2\right){\omega }^3=1\right] \\ =\omega \left[-1-\omega \right] \\ =\omega \left({\omega }^2\right)\left[\because -1-\omega ={\omega }^2\right] \\ ={\omega }^3 \\ =1\left[\mathrm{From}\left(2\right)\right] \end{gathered}$$

Hence, Option(A) is the correct answer.