Answer

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**Hint:**In the above expression, $\omega $ is the cube root of unity. And we know that a cube of $\omega $ is equal to 1 which will look as follows: ${{\omega }^{3}}=1$. Also, we know that sum of 1, $\omega $ and square of $\omega $ is equal to 0 and the mathematical expression for this addition will look as follows: $1+\omega +{{\omega }^{2}}=0$. Now, to simplify the above expression, we are going to write the powers of $\omega $ in the form of ${{\omega }^{3}}$ so that we can write and then simplify using basic algebra.

**Complete step-by-step solution:**

The expression given in the above problem is as follows:

$\left( 1-\omega \right)\left( 1-{{\omega }^{2}} \right)\left( 1-{{\omega }^{4}} \right)\left( 1-{{\omega }^{8}} \right)$

In the above expression, $\omega $ is the cube root of unity and the mathematical form of this cube root of unity is as follows:

$\omega ={{\left( 1 \right)}^{\dfrac{1}{3}}}$

Cubing both the sides of the above equation we get,

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

Also, there is a relation of the cube root of unity as follows:

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

Now, we are going to rearrange the given expression by writing ${{\omega }^{4}}={{\omega }^{3}}\left( \omega \right)$ and ${{\omega }^{8}}={{\omega }^{6}}\left( {{\omega }^{2}} \right)$ in the above expression and we get,

$\left( 1-\omega \right)\left( 1-{{\omega }^{2}} \right)\left( 1-{{\omega }^{3}}\left( \omega \right) \right)\left( 1-{{\omega }^{6}}\left( {{\omega }^{2}} \right) \right)$ ……… (1)

We have shown above that the cube of the cube root of unity is equal to 1.

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

Now, taking square on both the sides of the above equation we get,

$\begin{align}

& {{\left( {{\omega }^{3}} \right)}^{2}}={{\left( 1 \right)}^{2}} \\

& \Rightarrow {{\omega }^{6}}=1 \\

\end{align}$

Using the above relation in eq. (1) we get,

$\begin{align}

& \left( 1-\omega \right)\left( 1-{{\omega }^{2}} \right)\left( 1-\left( 1 \right)\left( \omega \right) \right)\left( 1-\left( 1 \right)\left( {{\omega }^{2}} \right) \right) \\

& =\left( 1-\omega \right)\left( 1-{{\omega }^{2}} \right)\left( 1-\omega \right)\left( 1-{{\omega }^{2}} \right) \\

& ={{\left( 1-\omega \right)}^{2}}{{\left( 1-{{\omega }^{2}} \right)}^{2}} \\

\end{align}$

Rearranging the powers in the above expression we get,

${{\left( \left( 1-\omega \right)\left( 1-{{\omega }^{2}} \right) \right)}^{2}}$

Multiplying $\left( 1-\omega \right)\And \left( 1-{{\omega }^{2}} \right)$ in the above expression and we get,

$\begin{align}

& {{\left( 1-{{\omega }^{2}}-\omega +{{\omega }^{3}} \right)}^{2}} \\

& ={{\left( 1-\left( {{\omega }^{2}}+\omega \right)+{{\omega }^{3}} \right)}^{2}}.......(3) \\

\end{align}$

In the above, we have shown that:

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

Subtracting 1 on both the sides we get,

$\omega +{{\omega }^{2}}=-1$

Using the above relation in eq. (3) we get,

$\begin{align}

& {{\left( 1-\left( -1 \right)+1 \right)}^{2}} \\

& ={{\left( 1+1+1 \right)}^{2}} \\

& ={{3}^{2}}=9 \\

\end{align}$

From the above solution, we have solved the given expression to 9.

**Hence, the simplification of the above expression is equal to 9.**

**Note:**To solve the above problem, you must know the properties of the cube root of unity otherwise you cannot solve this problem so make sure you have a good understanding of this concept of the cube root of unity. Also, don’t make any calculation mistakes in the above problem.

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