If $1,\ \omega,\ \omega^{2}$ are the cube roots of unity, then $\left( 1 - \omega + \omega^{2} \right)\left( 1 - \omega^{2} + \omega^{4} \right)\left( 1 - \omega^{4} + \omega^{8} \right)(1 - \omega^{8} + \omega^{16})$... upto $2n$ factors is

Question

Dr. Abubacker · Accepted Answer

(b)
$\left( 1 - \omega + \omega^{2} ight)\left( 1 - \omega^{2} + \omega^{3}.\omega ight)$
$\left( 1 - \omega^{3}.\omega + \omega^{6}.\omega^{2} ight)(1 - \omega^{6}.\omega^{2} + \omega^{15}.\omega)$… upto $2n$
$= \left( 1\omega + \omega^{2} ight)\left( 1 - \omega^{2} + \omega ight)$
$\left( 1 - \omega + \omega^{2} ight)(1 - \omega^{2} + \omega)$… upto $2n$
$= \left\lbrack ( - 2\omega)\left( - 2\omega^{2} ight) ightbrack 	imes \lbrack( - 2\omega)( - 2\omega^{2})brack 	imes$… upto $2n$
$= \left( 2^{2}\omega^{3} ight) 	imes (2^{2}\omega^{3}) 	imes$…. upto $n$
$=$[$2^{2} 	imes 2^{2} 	imes 2^{2}$×…upto n]$= 2^{2n}$

Q. If $1,\ \omega,\ \omega^{2}$ are the cube roots of unity, then $\left( 1 - \omega + \omega^{2} \right)\left( 1 - \omega^{2} + \omega^{4} \right)\left( 1 - \omega^{4} + \omega^{8} \right)(1 - \omega^{8} + \omega^{16})$... upto $2n$ factors is

Explanation

Reviewed by Dr. Abubacker

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