Question

If 1, w,w2 are cube roots of unity then evaluate (1 - w4 +w8) (1 - w8 + w16).​

Answer 1

$\large\underline{\sf{Solution-}}$

We know that

$\rm :\longmapsto\:1, \: \omega, \: { \omega}^{2} \: are \: cube \: roots \: of \: unity \: then$

$\boxed{ \bf{ \: 1 + \omega \: + { \omega}^{2} = 0}}$

$\boxed{ \bf{ \: 1 + \omega = - { \omega}^{2}}}$

$\boxed{ \bf{ \: 1 + { \omega}^{2} = \omega}}$

$\boxed{ \bf{ \: \omega + { \omega}^{2} = - 1}}$

$\boxed{ \bf{ \: { \omega}^{3} = 1}}$

Now, Consider

$\red{\rm :\longmapsto\:(1 - { \omega}^{4} + { \omega}^{8})(1 - { \omega}^{8} + { \omega}^{16})}$

can be rewritten as

$\rm \:  =(1 - { \omega}^{3}. \omega + { \omega}^{6}. { \omega}^{2})(1 - { \omega}^{6}. { \omega}^{2} + { \omega}^{15}. \omega)$

We know,

$\boxed{ \bf{ \: { \omega}^{3} = { \omega}^{6} = { \omega}^{15} = 1}}$

So, using this, we get

$\rm \:  =  \:  \: (1 - \omega + { \omega}^{2} )(1 - { \omega}^{2} + \omega)$

can be re-arranged as

$\rm \:  =  \:  \: (1 + { \omega}^{2} - \omega)(1 + \omega - { \omega}^{2})$

$\rm \:  =  \:  \: ( - \omega - \omega)( - { \omega}^{2} - { \omega}^{2})$

$\rm \:  =  \:  \: ( - 2 \omega)( - 2 { \omega}^{2})$

$\rm \:  =  \:  \: {4 \omega}^{3}$

$\rm \:  =  \:  \: 4 \times 1$

$\rm \:  =  \:  \: 4$

Hence,

$\red{\bf :\longmapsto\:(1 - { \omega}^{4} + { \omega}^{8})(1 - { \omega}^{8} + { \omega}^{16}) = 4}$

Additional Information :-

Cube roots of unity

$\rm :\longmapsto\:x = {\bigg(1\bigg) }^{\dfrac{1}{3} }$

On cubing both sides, we get

$\rm :\longmapsto\: {x}^{3} = 1$

$\rm :\longmapsto\: {x}^{3} - 1 = 0$

$\rm :\longmapsto\:(x - 1)( {x}^{2} + x + 1) = 0$

$\bf\implies \:x = 1$

or

$\bf\implies \: {x}^{2} + x + 1 = 0$

So, using Quadratic formula, we get

$\rm :\longmapsto\:x = \dfrac{ - 1 \: \pm \: \sqrt{ {1}^{2} - 4 \times 1 \times 1 } }{2 \times 1}$

$\rm :\longmapsto\:x = \dfrac{ - 1 \: \pm \: \sqrt{ 1 - 4} }{2 }$

$\rm :\longmapsto\:x = \dfrac{ - 1 \: \pm \: \sqrt{ - 3} }{2 }$

$\rm :\longmapsto\:x = \dfrac{ - 1 \: \pm \: i \: \sqrt{3} }{2 }$

So, it means, x have the values

$\rm :\longmapsto\:x = \: 1 \: \: or \: \dfrac{ - 1 \: + \: i \: \sqrt{3} }{2 } \: \: or \: \: \dfrac{ - 1 \: - \: i \: \sqrt{3} }{2 }$

So, here,

$\rm :\longmapsto\: \omega = \dfrac{ - 1 \: + \: i \: \sqrt{3} }{2 }$

and

$\rm :\longmapsto\: { \omega}^{2} = \dfrac{ - 1 \: - \: i \: \sqrt{3} }{2 }$