If $1,\ \omega,\ \omega^{2}$ are the cube roots of unity, then $(1 + \omega)\left( 1 + \omega^{2} \right)\left( 1 + \omega^{4} \right)(1 + \omega^{8})$ is equal to

Question

Dr. Abubacker · Accepted Answer

(a)
Given, $(1 + \omega)\left( 1 + \omega^{2} \right)\left( 1 + \omega^{4} \right)(1 + \omega^{8})$
$= (1 + \omega)( - \omega)(1 + \omega)\left( 1 + \omega^{2} \right)$
[$\because$1+$\ \omega + \omega^{2} = 0$ and $\omega^{4} = \omega\rbrack$
$= (1 + \omega)^{2}( - \omega - \omega^{3})$
$= \left( 1 + \omega^{2} + 2\omega \right)( - \omega - 1)$
$= (\omega)\left( \omega^{2} \right) = 1$

Q. If $1,\ \omega,\ \omega^{2}$ are the cube roots of unity, then $(1 + \omega)\left( 1 + \omega^{2} \right)\left( 1 + \omega^{4} \right)(1 + \omega^{8})$ is equal to

Explanation

Reviewed by Dr. Abubacker

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