Question

If $\omega$ is a cube root of unity then the value of $(1-\omega +{\omega }^2{)}^5+(1+\omega -{\omega }^2{)}^5$ is

• A. $30$
• B. $32$
• C. $2$
• D. None of these

Answer 1

The correct option is B

$32$

Explanation for the correct option

Step 1: Prerequisites for the solution

Given that $\omega$ is a complex cube root of unity then we know that

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

From this we have

$$\begin{gathered} 1+{\omega }^2=-\omega \dots \left(1\right) \\ 1+\omega =-{\omega }^2\dots \left(2\right) \end{gathered}$$

Step 2: Calculation for the correct option

As the given expression is $(1-\omega +{\omega }^2{)}^5+(1+\omega -{\omega }^2{)}^5$ then from the equation $\left(1\right)\text{and}\left(2\right)$ we have,

$\begin{array}{rcl}\therefore {\left(1-\omega +{\omega }^2\right)}^5+{\left(1+\omega -{\omega }^2\right)}^5& =& {\left(-\omega -\omega \right)}^5+{\left(-{\omega }^2-{\omega }^2\right)}^5\\ & =& {\left(-2\omega \right)}^5+{\left(-2{\omega }^2\right)}^5\\ & =& -32{\omega }^5-32{\omega }^{10}\\ & =& -32\left({\omega }^3·{\omega }^2\right)-32\left({\left({\omega }^3\right)}^3·\omega \right)\\ & =& -32\left({\omega }^2+\omega \right)\\ & =& -32\left(-1\right)\\ & =& 32\end{array}$

Hence, the correct option is (B).