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

$1$

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B

$-1$

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C

$i$

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D

$0$

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Solution

## 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.$1+\omega +{\omega }^{2}=0___\left(1\right)\left[\mathbf{\because }\mathbf{Using}\mathbf{}\mathbf{definition}\mathbf{,}\mathbf{}\mathbf{}\mathbf{\omega }\mathbf{=}\frac{\mathbf{-}\mathbf{1}\mathbf{+}\mathbf{i}\sqrt{\mathbf{3}}}{\mathbf{2}}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{\omega }}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{-}\mathbf{1}\mathbf{-}\mathbf{i}\sqrt{\mathbf{3}}}{\mathbf{2}}\right]\phantom{\rule{0ex}{0ex}}1×\mathrm{\omega }×{\mathrm{\omega }}^{2}=1\mathrm{____}\left(2\right)$Step2. Find the value of $\mathbf{\omega }\left({\left(1+\omega \right)}^{3}-{\left(1+\omega \right)}^{2}\right)$:$\omega \left({\left(1+\omega \right)}^{3}-{\left(1+\omega \right)}^{2}\right)\phantom{\rule{0ex}{0ex}}=\omega \left({\left(-{\omega }^{2}\right)}^{3}-{\left(-{\omega }^{2}\right)}^{2}\right)\left[\because 1+\omega =-{\omega }^{2}\right]\phantom{\rule{0ex}{0ex}}=\omega \left[{\left(-1\right)}^{3}{\left({\omega }^{3}\right)}^{2}-\left({\omega }^{3}\right)\left(\omega \right)\right]\phantom{\rule{0ex}{0ex}}=\omega \left[-1{\left(1\right)}^{2}-\left(1\right)\left(\omega \right)\right]\left[\because From\left(2\right){\omega }^{3}=1\right]\phantom{\rule{0ex}{0ex}}=\omega \left[-1-\omega \right]\phantom{\rule{0ex}{0ex}}=\omega \left({\omega }^{2}\right)\left[\because -1-\omega ={\omega }^{2}\right]\phantom{\rule{0ex}{0ex}}={\omega }^{3}\phantom{\rule{0ex}{0ex}}=1\left[\mathrm{From}\left(2\right)\right]$Hence, Option(A) is the correct answer.

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