Question

cube roots of unity 2 If w & we are О Prove that (2-w+2 w²) (2+2w - w^2) = 9​

Answer 1

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

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

$\large\underline{\sf{To\:prove- }}$

$\rm :\longmapsto\:(2 - \omega + 2 {\omega }^{2})(2 + 2\omega - {\omega }^{2}) = 9$

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

We know,

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

then

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

and

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

Now, Consider,

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

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

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

$\rm \: = \: 2( -\omega ) - \omega$

$\rm \: = \: - 2\omega - \omega$

$\rm \: = \: - 3\omega$

Hence,

$\rm :\implies\:\purple{ \boxed{ \bf{2 - \omega + 2 {\omega }^{2} = - 3\omega }}}$

Now, Consider

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

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

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

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

$\rm \: = \: { - 3\omega }^{2}$

Hence,

$\rm :\implies\:\purple{ \boxed{ \bf{ \: \: 2 + 2\omega - {\omega }^{2} = - {3\omega }^{2} }}}$

Now, Consider

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

On substituting the values evaluated above, we have

$\rm \: = \: ( - 3\omega )( - {3\omega }^{2} )$

$\rm \: = \: {9\omega }^{3}$

$\rm \: = \: 9$

Hence,

$\bf\implies \:\purple{ \boxed{ \bf{(2 - \omega + 2 {\omega }^{2})(2 + 2\omega - {\omega }^{2}) = 9}}}$

Additional Information :-

Let derive cube roots of unity!!!

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

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

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

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

$\bf\implies \:x = 1$

or

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

Its a quadratic equation in x, so using Quadratic Formula we get

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

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

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

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

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

Hence,

Cube roots of unity are

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

represented as

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