Question

Find cube root of unity and hence proved 1+w+w^=0

Answer 1

Here is the attached solution of your problem.

Answer 2

Appropriate Question :-

Find the cube roots of unity and hence proved that

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

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

Let assume that

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

On cubing both sides, we get

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

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

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

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

$\: \: \: \: [ \: \because \: \rm \: {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2})]$

$\rm\implies \:x = 1$

or

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

So, its a quadratic equation in x, so we have to use Quadratic Formula to get the values of x.

So, by using Quadratic formula,

$\boxed{\sf{  \: \: x \: = \: \frac{ - \: b \: \pm \: \sqrt{ {b}^{2} - 4ac} }{2a} \: }}$

So, here

$\rm \:\: \: \: \: \: a = 1$

$\rm \:\: \: \: \: \: b = 1$

$\rm \:\: \: \: \: \: c = 1$

So, on substituting the values, we get

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

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

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

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

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

$\rm\implies \:x = \omega \: or \: \: {\omega }^{2}$

where,

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

and

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

Hence, Cube roots of unity are

$\rm\implies \:\boxed{\sf{  \: \: \rm \: {\bigg( 1\bigg) }^{\dfrac{1}{3} } = 1, \: \omega , \: {\omega }^{2} \: \: }}$

Now, Consider

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

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

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

$\rm \: =  \: \dfrac{0}{2}$

$\rm \: =  \: 0$

Hence,

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

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Additional Information :-

$\begin{gathered}\begin{gathered}\bf\: 1 + {\omega }^{n} + {\omega }^{2n} = \begin{cases} &\sf{0 \: if \: n \: is \: not \: a \: multiple \: of \: 3} \\ \\ &\sf{3 \: if \: n \: is \: a \: multiple \: of \: 3} \end{cases}\end{gathered}\end{gathered}$

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