Question

Let $\omega $ being the cube root of unity then find the value of ${{\left( 1+\omega +{{\omega }^{2}} \right)}^{5}}$ .

Answer 1

Hint: Use the property that ${{\omega }^{3}}=1$ and $1+\omega +{{\omega }^{2}}=0$ , where $\omega $ represents cube root of unity.

Complete step-by-step answer:

Before starting with the solution, let us discuss some of the formulas related to the cube root of unity. We should know that the cube root of unity can be mathematically written in the form of the equation as:

${{x}^{3}}=1$

This can be further simplified to the polynomial form as:

${{x}^{3}}-1=0$

The important formulas include:

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

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

Now let us start with the solution to the expression given in the question.

${{\left( 1+\omega +{{\omega }^{2}} \right)}^{5}}$

When we use the property $1+\omega +{{\omega }^{2}}=0$ , we get

 ${{\left( 1+\omega +{{\omega }^{2}} \right)}^{5}}$

$={{0}^{5}}$

Now we know that 0 to the power any positive finite number is equal to zero, and we know that 5 is positive, as well as finite. So, we can conclude that the value of ${{\left( 1+\omega +{{\omega }^{2}} \right)}^{5}}$ is 0 provided $\omega $ represents cube root of unity.


Note: Don’t get confused and apply the properties of the cube root of unity to all the places wherever you see the symbol $\omega $ , as the symbol has different meanings in different chapters and topics. So, be sure that you use the above mentioned properties with $\omega $ only if it is mentioned that $\omega $ represents the cube root of unity. Also, be careful while opening the brackets and multiplying the signs. Often, students commit mistakes while opening the brackets.