Question

prove (2+ω+ω^2)^3 +(1+ω-ω^2)^8 -(1-3ω+ω^2)^4 =1, where 'ω' is omega in cube roots of unity

Answer 1

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

Consider,

$\rm :\longmapsto\: {(2 + \omega + {\omega }^{2} )}^{3} + {(1 + \omega - {\omega }^{2})}^{8} - {(1 - 3\omega + {\omega }^{2})}^{4}$

can be re-arranged as

$\rm = {\bigg[2 + (\omega + {\omega }^{2} )\bigg]}^{3} + {\bigg[(1 + \omega ) - {\omega }^{2} \bigg]}^{8} - {\bigg[(1 + {\omega }^{2}) - 3\omega \bigg]}^{4}$

We know,

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

So, using this, we get

$\rm = {\bigg[2 + ( - 1)\bigg]}^{3} + {\bigg[ - {\omega }^{2} - {\omega }^{2} \bigg]}^{8} - {\bigg[( - \omega ) - 3\omega \bigg]}^{4}$

$\rm = {\bigg[2 - 1\bigg]}^{3} + {\bigg[ -2 {\omega }^{2}\bigg]}^{8} - {\bigg[ - 4\omega \bigg]}^{4}$

$\rm = {\bigg[1\bigg]}^{3} + {\bigg[ 2 {\omega }^{2}\bigg]}^{8} - {\bigg[ 4\omega \bigg]}^{4}$

$\rm \:  =  \: 1 + 256 {\omega }^{16} - 256 {\omega }^{4}$

$\rm \:  =  \: 1 + 256 {\omega }^{15} \times\omega - 256 {\omega }^{3} \times \omega$

We know,

$\red{ \boxed{ \sf{ \: {\omega }^{3} = 1 \: }}}$

So, using this identity, we get

$\rm \:  =  \: 1 + 256 \omega - 256 \omega$

$\rm \:  =  \: 1$

Hence, Proved

Additional Information :-

Proof of Complex 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

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

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

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

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

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

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

$\bf\implies \:x = 1, \: \dfrac{ - 1 \: + \: i \: \sqrt{3} }{2} , \: \dfrac{ - 1 \: - \: i \: \sqrt{3} }{2}$

or

$\bf\implies \:x = 1, \omega , \: {\omega }^{2}$

where,

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

and

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

So,

$\rm :\longmapsto\:1 + \omega + {\omega }^{2}$

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

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

$\rm \:  =  \: 0$