Question

If 1,w,w2 are the three cube roots of unity then (3+w2+w4)6= ?​

Answer 1

Step-by-step explanation:

$(3 + {\omega}^{2} + {\omega}^{4} ) ^{6}$

we know that

$1 + \omega + {\omega}^{2} = 0 \\ and \: \: \\ {\omega}^{3} = 1$

$(3 + {\omega}^{2} + {\omega}^{4} ) ^{6} \\ = (2 + 1+ {\omega}^{2} + {\omega}^{3}.\omega ) ^{6} \\ = (2 +1 + \omega + {\omega}^{2} ) {}^{6} \\ = (2 + 0) {}^{6} \\ = {2}^{6} \\ = 64$

Answer 2

Given:

$1,\omega,\omega^{2}$ are the three cube roots of unity.

To find:

The value of $=(2+1+\omega^{2}+\omega^{3}.\omega)$.

Solution:

We need to know that,

$\omega^{3}=1...(1)$

and, $1+\omega+\omega^{2}=0$

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

Here, we have $(3+\omega^{2}+\omega^{4})6$ . This should be split in the form of $1,\omega,\omega^{2}$so that we can write the expression in the form of equations (1) and (2) for easier simplification:

$(3+\omega^{2}+\omega^{4})6$

$3$ is split up as $2+1$ and $\omega^{4}$ is split as $\omega^{3}.\omega$.

Hence,

$=(2+1+\omega^{2}+\omega^{3}.\omega)6$

Substituting $1+\omega^{2}=-\omega$ and $\omega^{3}=1$ ,

$=(2+-\omega+\omega)6$

Positive and negative terms of the same value will be cancelled.

$=(2)6=12$

∴ $(3+\omega^{2}+\omega^{4})6=12$

If $1,\omega,\omega^{2}$ are the three cube roots of unity, then $(3+\omega^{2}+\omega^{4})6=12$.