Q

# BITSAT If 1, w, w^2 are the cube roots of unity, the the value of

(1+$\omega^{2}$-$\omega$)(1-$\omega^{2}$+$\omega$) is

1. 4

2. $\omega$

3. 2

4. Zero

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With the help of 2 properties of cube roots of unity, we can solve this question.

Property 1, $\omega ^{3} = 1$

Property 2, $1+ \omega +\omega ^{2} = 0$

Now we have to find the value of (1+$\omega^{2}$-$\omega$)(1-$\omega^{2}$+$\omega$)

Using property 2, $(-\omega -\omega )(-\omega ^{2} - \omega ^{2}) = (-2\omega )(-2\omega ^{2}) = 4\omega ^{3}$

Now use property 1,

$4\omega ^{3} = 4$

Option (1) is correct

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