# Question #3bc5d

Dec 16, 2016

The answer is $16 \omega$.

#### Explanation:

Since, $\sqrt[3]{1} = \omega$,
So, ${\omega}^{3} = 1$.-----(1).

$\therefore {\omega}^{3} - 1 = 0$
$\therefore \left(\omega - 1\right) \left({\omega}^{2} + \omega + 1\right) = 0$

So, either first term is zero, or the second term is zero.

But, if $\omega - 1 = 0$, $\omega = 1$, which is not true.

So, ${\omega}^{2} + \omega + 1 = 0$.
$\therefore {\omega}^{2} + 1 = - \omega$.----(2).

Now, ${\left(3 + \omega + 3 {\omega}^{2}\right)}^{4}$
$= {\left\{\omega + 3 \left({\omega}^{2} + 1\right)\right\}}^{4}$
$= {\left\{\omega + 3 \left(- \omega\right)\right\}}^{4}$ ----[From (2)].
$= {\left(\omega - 3 \omega\right)}^{4}$
$= {\left(- 2 \omega\right)}^{4}$
$= 16 {\omega}^{4}$
$= 16 {\omega}^{3} \times \omega$
$= 16 \times 1 \times \omega$ ----[From (1)].
$= 16 \omega$. (Answer).

Hope it Helps :)