Question

The value of ${{\left( 1+{{\omega }^{2}}+2\omega \right)}^{3n}}-{{\left( 1+\omega +2{{\omega }^{2}} \right)}^{3n}}$ is equal to?
$\text{A) }Zero$
$\text{B) }1$
$\text{C) }\omega $
$\text{D) }{{\omega }^{2}}$

Answer 1

Hint: In this question we have been an equation which has the term $\omega $ in it which is the cube root of negative unity which is represented as $\omega =\dfrac{-1\pm i\sqrt{3}}{2}$. In this question we will use the property $1+\omega +{{\omega }^{2}}=0$ and ${{\omega }^{3}}=1$by first converting the parts of the expression into this format and then simplify the expression to get the required answer.

Complete step by step solution:
We have the expression given to us as:
$\Rightarrow {{\left( 1+{{\omega }^{2}}+2\omega \right)}^{3n}}-{{\left( 1+\omega +2{{\omega }^{2}} \right)}^{3n}}$
Now consider $A={{\left( 1+{{\omega }^{2}}+2\omega \right)}^{3n}}$ and $B={{\left( 1+\omega +2{{\omega }^{2}} \right)}^{3n}}$ therefore the expression becomes $A-B$.
Now consider $A$,
$\Rightarrow {{\left( 1+{{\omega }^{2}}+2\omega \right)}^{3n}}$
In the expression, we have the term $2\omega $, on splitting the term, we get:
$\Rightarrow {{\left( 1+{{\omega }^{2}}+\left( \omega +\omega \right) \right)}^{3n}}$
On simplifying the bracket, we get:
$\Rightarrow {{\left( 1+{{\omega }^{2}}+\omega +\omega \right)}^{3n}}$
On rearranging the expression, we get:
$\Rightarrow {{\left( 1+\omega +{{\omega }^{2}}+\omega \right)}^{3n}}$
Now we have the first three terms of the expression in the form of $1+\omega +{{\omega }^{2}}$ and by using the property that $1+\omega +{{\omega }^{2}}=0$, we get:
$\Rightarrow {{\left( 0+\omega \right)}^{3n}}$
On simplifying, we get:
$\Rightarrow A={{\left( \omega \right)}^{3n}}$
Now consider $B$,
$\Rightarrow \left( 1+\omega +2{{\omega }^{2}} \right)$
In the expression, we have the term $2{{\omega }^{2}}$, on splitting the term, we get:
$\Rightarrow {{\left( 1+\omega +\left( {{\omega }^{2}}+{{\omega }^{2}} \right) \right)}^{3n}}$
On simplifying the bracket, we get:
$\Rightarrow {{\left( 1+\omega +{{\omega }^{2}}+{{\omega }^{2}} \right)}^{3n}}$
Now we have the first three terms of the expression in the form of $1+\omega +{{\omega }^{2}}$ and by using the property that $1+\omega +{{\omega }^{2}}=0$, we get:
$\Rightarrow {{\left( 0+{{\omega }^{2}} \right)}^{3n}}$
On simplifying, we get:
$\Rightarrow B={{\left( {{\omega }^{2}} \right)}^{3n}}$
Now the expression is in the form of $A-B$, on substituting the values, we get:
$\Rightarrow {{\omega }^{3n}}-{{\left( {{\omega }^{2}} \right)}^{3n}}$
Now we know the property of exponents that ${{a}^{bc}}={{\left( {{a}^{b}} \right)}^{c}}$, on using this property, we get:
$\Rightarrow {{\left( {{\omega }^{3}} \right)}^{n}}-{{\left( {{\omega }^{2}} \right)}^{3n}}$
Now we know the property of exponents that ${{\left( {{a}^{b}} \right)}^{cd}}={{\left( {{a}^{c}} \right)}^{bd}}$, on using this property, we get:
$\Rightarrow {{\left( {{\omega }^{3}} \right)}^{n}}-{{\left( {{\omega }^{3}} \right)}^{2n}}$
Now we know that ${{\omega }^{3}}=1$, on substituting, we get:
$\Rightarrow {{\left( 1 \right)}^{n}}-{{\left( 1 \right)}^{2n}}$
Now we know that $1$ raised to any real number is $1$, therefore, we get:
$\Rightarrow 1-1$
On simplifying, we get:
$\Rightarrow 0$, which is the required solution.

So, the correct answer is “Option A”.

Note: It is to be remembered that $\omega $ is used in complex numbers for ease in factorization. $\omega $ is the root of the quadratic equation ${{x}^{2}}+x+1=0$. The term $i$ in the value of $\omega $ represents the complex number which has a value $\sqrt{-1}$ also referred to as the square root of negative $1$.